RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.2

Question 1. In the following expansions, find the term as stated:
(i) 5th term of \( (a + 2x^3)^{17} \)
(ii) 9th term of \( {\left(\frac {x}{ y } -\frac { 3y }{{x}^{2}} \right) }^{12} \)
(iii) 6th term of \( {\left(\frac {2}{\sqrt {x}}-\frac{{x}^{2}}{ 2 } \right) }^{9} \)
Answer:
(i) For the 5th term of \( (a + 2x^3)^{17} \), we use the general formula for the \( (r+1)^{th} \) term in the expansion of \( (A+B)^n \), which is \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = a \), \( B = 2x^3 \), \( n = 17 \).
Since we need the 5th term, \( r+1 = 5 \), which means \( r = 4 \).
So, the 5th term \( T_5 \) is:
\( T_5 = T_{4+1} = ^{17}C_4 (a)^{17-4} (2x^3)^4 \)
\( T_5 = ^{17}C_4 a^{13} (2^4 (x^3)^4) \)
\( T_5 = ^{17}C_4 a^{13} (16x^{12}) \)
\( T_5 = 1820 \times 16 a^{13}x^{12} \)
\( T_5 = 29120 a^{13}x^{12} \)
(ii) For the 9th term of \( {\left(\frac {x}{ y } -\frac { 3y }{{x}^{2}} \right) }^{12} \), we use the same formula.
Here, \( A = \frac{x}{y} \), \( B = -\frac{3y}{x^2} \), \( n = 12 \).
Since we need the 9th term, \( r+1 = 9 \), which means \( r = 8 \).
So, the 9th term \( T_9 \) is:
\( T_9 = T_{8+1} = ^{12}C_8 {\left(\frac{x}{y}\right)}^{12-8} {\left(-\frac{3y}{x^2}\right)}^8 \)
\( T_9 = ^{12}C_8 {\left(\frac{x}{y}\right)}^4 {\left(-\frac{3y}{x^2}\right)}^8 \)
\( T_9 = 495 \times \frac{x^4}{y^4} \times \frac{(3y)^8}{(x^2)^8} \)
\( T_9 = 495 \times \frac{x^4}{y^4} \times \frac{3^8 y^8}{x^{16}} \)
\( T_9 = 495 \times \frac{3^8 y^{8-4}}{x^{16-4}} \)
\( T_9 = 495 \times \frac{6561 y^4}{x^{12}} \)
\( T_9 = \frac{3247695 y^4}{x^{12}} \)
(iii) For the 6th term of \( {\left(\frac {2}{\sqrt {x}}-\frac{{x}^{2}}{ 2 } \right) }^{9} \), we use the same formula.
Here, \( A = \frac{2}{\sqrt{x}} = 2x^{-1/2} \), \( B = -\frac{x^2}{2} \), \( n = 9 \).
Since we need the 6th term, \( r+1 = 6 \), which means \( r = 5 \).
So, the 6th term \( T_6 \) is:
\( T_6 = T_{5+1} = ^9C_5 {\left(2x^{-1/2}\right)}^{9-5} {\left(-\frac{x^2}{2}\right)}^5 \)
\( T_6 = ^9C_5 {\left(2x^{-1/2}\right)}^4 {\left(-\frac{x^2}{2}\right)}^5 \)
\( T_6 = 126 \times (2^4 x^{-4/2}) \times \left(-\frac{x^{10}}{2^5}\right) \)
\( T_6 = 126 \times (16 x^{-2}) \times \left(-\frac{x^{10}}{32}\right) \)
\( T_6 = 126 \times \frac{16}{32} \times (-1) x^{10-2} \)
\( T_6 = 126 \times \frac{1}{2} \times (-1) x^8 \)
\( T_6 = -63 x^8 \)
In simple words: To find a specific term in a binomial expansion, use the general formula \( T_{r+1} = nC_r A^{n-r} B^r \). Plug in the values for 'A', 'B', 'n', and 'r' (which is one less than the term number you want). Then, simplify the expression to get the final term.

🎯 Exam Tip: Remember to correctly identify 'A', 'B', and 'n' from the binomial expression and 'r' from the term number. Pay close attention to signs and exponents when simplifying.

Question 2. Find the coefficient of the stated power of x in the following expansions:
(i) \( x^{-7} \) in the expansion of \( {\left(ax-\frac { 1 }{ { bx }^{2}} \right) }^{8} \)
(ii) \( x^4 \) in the expansion of \( {\left( {x}^{4}+\frac {1}{{x}^{3}} \right)}^{15} \)
(iii) \( x^6 \) in the expansion \( (a – bx^2)^{10} \)
Answer:
(i) To find the coefficient of \( x^{-7} \) in the expansion of \( {\left(ax-\frac { 1 }{ { bx }^{2}} \right) }^{8} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = ax \), \( B = -\frac{1}{bx^2} = -b^{-1}x^{-2} \), \( n = 8 \).
So, the \( (r+1)^{th} \) term is:
\( T_{r+1} = ^8C_r (ax)^{8-r} {\left(-\frac{1}{bx^2}\right)}^r \)
\( T_{r+1} = ^8C_r a^{8-r} x^{8-r} (-1)^r (b^{-1})^r (x^{-2})^r \)
\( T_{r+1} = (-1)^r ^8C_r a^{8-r} b^{-r} x^{8-r-2r} \)
\( T_{r+1} = (-1)^r ^8C_r a^{8-r} b^{-r} x^{8-3r} \)
For the coefficient of \( x^{-7} \), the power of \( x \) should be \( -7 \).
So, \( 8 - 3r = -7 \)
\( -3r = -7 - 8 \)
\( -3r = -15 \)
\( r = 5 \)
Now, substitute \( r=5 \) back into the coefficient part of \( T_{r+1} \):
Coefficient of \( x^{-7} = (-1)^5 ^8C_5 a^{8-5} b^{-5} \)
Coefficient of \( x^{-7} = (-1) \times 56 \times a^3 b^{-5} \)
Coefficient of \( x^{-7} = -\frac{56a^3}{b^5} \)

(ii) To find the coefficient of \( x^4 \) in the expansion of \( {\left( {x}^{4}+\frac {1}{{x}^{3}} \right)}^{15} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = x^4 \), \( B = \frac{1}{x^3} = x^{-3} \), \( n = 15 \).
So, the \( (r+1)^{th} \) term is:
\( T_{r+1} = ^{15}C_r (x^4)^{15-r} (x^{-3})^r \)
\( T_{r+1} = ^{15}C_r x^{4(15-r)} x^{-3r} \)
\( T_{r+1} = ^{15}C_r x^{60-4r-3r} \)
\( T_{r+1} = ^{15}C_r x^{60-7r} \)
For the coefficient of \( x^4 \), the power of \( x \) should be \( 4 \).
So, \( 60 - 7r = 4 \)
\( -7r = 4 - 60 \)
\( -7r = -56 \)
\( r = 8 \)
Now, substitute \( r=8 \) back into the coefficient part of \( T_{r+1} \):
Coefficient of \( x^4 = ^{15}C_8 \)
\( ^{15}C_8 = \frac{15!}{8!(15-8)!} = \frac{15!}{8!7!} \)
\( ^{15}C_8 = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( ^{15}C_8 = 15 \times 13 \times 11 \times 3 = 6435 \)

(iii) To find the coefficient of \( x^6 \) in the expansion of \( (a – bx^2)^{10} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = a \), \( B = -bx^2 \), \( n = 10 \).
So, the \( (r+1)^{th} \) term is:
\( T_{r+1} = ^{10}C_r (a)^{10-r} (-bx^2)^r \)
\( T_{r+1} = ^{10}C_r a^{10-r} (-1)^r b^r (x^2)^r \)
\( T_{r+1} = (-1)^r ^{10}C_r a^{10-r} b^r x^{2r} \)
For the coefficient of \( x^6 \), the power of \( x \) should be \( 6 \).
So, \( 2r = 6 \)
\( r = 3 \)
Now, substitute \( r=3 \) back into the coefficient part of \( T_{r+1} \):
Coefficient of \( x^6 = (-1)^3 ^{10}C_3 a^{10-3} b^3 \)
Coefficient of \( x^6 = (-1) \times \frac{10!}{3!7!} \times a^7 b^3 \)
Coefficient of \( x^6 = (-1) \times \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times a^7 b^3 \)
Coefficient of \( x^6 = -1 \times 120 \times a^7 b^3 \)
Coefficient of \( x^6 = -120 a^7 b^3 \)
In simple words: To find the coefficient of a specific power of x in an expansion, first write the general term \( T_{r+1} \) using the binomial formula. Then, set the exponent of x in the general term equal to the desired power and solve for 'r'. Finally, plug this 'r' value back into the coefficient part of the general term to get your answer.

🎯 Exam Tip: When finding coefficients, ensure you correctly handle negative signs and exponents for both the base terms and the variable 'x'. Mistakes in calculating 'r' or simplifying the combinatorial part are common.

Question 3. In the following expansions find the term independent of x:
(i) \( {\left(\frac {x}{2}+2y \right)}^{6} \)
(ii) \( {\left(3a-\frac{a^3}{6} \right)}^{9} \)
(iii) \( {\left(\sqrt {\frac {x}{3}}+\frac { 3 }{2{x}^{ 2 } } \right) }^{10} \)
(iv) \( {\left(3x-\frac{2}{x^2} \right)}^{15} \)
Answer:
(i) To find the term independent of x in \( {\left(\frac {x}{2}+2y \right)}^{6} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = \frac{x}{2} \), \( B = 2y \), \( n = 6 \).
\( T_{r+1} = ^6C_r {\left(\frac{x}{2}\right)}^{6-r} (2y)^r \)
\( T_{r+1} = ^6C_r \frac{x^{6-r}}{2^{6-r}} 2^r y^r \)
\( T_{r+1} = ^6C_r x^{6-r} 2^{r-(6-r)} y^r \)
\( T_{r+1} = ^6C_r x^{6-r} 2^{2r-6} y^r \)
For the term independent of x, the power of x should be 0.
So, \( 6 - r = 0 \)
\( r = 6 \)
Substitute \( r=6 \) into the term:
Term independent of \( x = ^6C_6 (2^{2(6)-6}) y^6 \)
\( = 1 \times 2^{12-6} y^6 \)
\( = 1 \times 2^6 y^6 \)
\( = 64 y^6 \)

(ii) To find the term independent of x in \( {\left(3a-\frac{a^3}{6} \right)}^{9} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
This expression has no 'x' variable. Therefore, every term in this expansion is independent of x. The question asks for "the term independent of x", which implies a specific value or constant. If there's no 'x' variable in the expression, then all terms are independent of 'x'. If the intention was to find the constant term (i.e., a term with no variable at all), the structure of the question usually implies a variable that cancels out. Since there is no 'x' in this expression, the expression itself is already independent of x.
The expansion is for \( (3a - \frac{a^3}{6})^9 \). Let's assume the question meant "term independent of a". If it were independent of 'a', then 'a' would need to be treated as 'x'. But it specifically asks for 'x'.
Given the input, the problem statement seems to have a mismatch, as there is no 'x' in the expression \( {\left(3a-\frac{a^3}{6} \right)}^{9} \). Thus, all terms are independent of x.
However, if the question implicitly meant 'a' as the variable, it would be a complex calculation to find the term where 'a' power becomes 0. Assuming the problem is strictly about 'x', there is no x to be independent of.
Let's assume there's a typo in the question and it should have contained 'x'. Without it, any term is independent of x.

(iii) To find the term independent of x in \( {\left(\sqrt {\frac {x}{3}}+\frac { 3 }{2{x}^{ 2 } } \right) }^{10} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = \sqrt{\frac{x}{3}} = x^{1/2} 3^{-1/2} \), \( B = \frac{3}{2x^2} = 3 \cdot 2^{-1} x^{-2} \), \( n = 10 \).
\( T_{r+1} = ^{10}C_r (x^{1/2} 3^{-1/2})^{10-r} (3 \cdot 2^{-1} x^{-2})^r \)
\( T_{r+1} = ^{10}C_r x^{(10-r)/2} 3^{-(10-r)/2} 3^r 2^{-r} x^{-2r} \)
\( T_{r+1} = ^{10}C_r 3^{r - (10-r)/2} 2^{-r} x^{(10-r)/2 - 2r} \)
For the term independent of x, the power of x should be 0.
So, \( \frac{10-r}{2} - 2r = 0 \)
\( 10 - r - 4r = 0 \)
\( 10 - 5r = 0 \)
\( 5r = 10 \)
\( r = 2 \)
Substitute \( r=2 \) into the term:
Term independent of \( x = ^{10}C_2 3^{2 - (10-2)/2} 2^{-2} \)
\( = ^{10}C_2 3^{2 - 8/2} 2^{-2} \)
\( = ^{10}C_2 3^{2 - 4} 2^{-2} \)
\( = ^{10}C_2 3^{-2} 2^{-2} \)
\( = \frac{10 \times 9}{2 \times 1} \times \frac{1}{3^2} \times \frac{1}{2^2} \)
\( = 45 \times \frac{1}{9} \times \frac{1}{4} \)
\( = \frac{45}{36} = \frac{5}{4} \)

(iv) To find the term independent of x in \( {\left(3x-\frac{2}{x^2} \right)}^{15} \), we use the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \).
Here, \( A = 3x \), \( B = -\frac{2}{x^2} = -2x^{-2} \), \( n = 15 \).
\( T_{r+1} = ^{15}C_r (3x)^{15-r} (-2x^{-2})^r \)
\( T_{r+1} = ^{15}C_r 3^{15-r} x^{15-r} (-2)^r x^{-2r} \)
\( T_{r+1} = ^{15}C_r 3^{15-r} (-2)^r x^{15-r-2r} \)
\( T_{r+1} = ^{15}C_r 3^{15-r} (-2)^r x^{15-3r} \)
For the term independent of x, the power of x should be 0.
So, \( 15 - 3r = 0 \)
\( 3r = 15 \)
\( r = 5 \)
Substitute \( r=5 \) into the term:
Term independent of \( x = ^{15}C_5 (3)^{15-5} (-2)^5 \)
\( = ^{15}C_5 (3)^{10} (-32) \)
\( = \frac{15!}{5!10!} \times 3^{10} \times (-32) \)
\( = 3003 \times 59049 \times (-32) \)
\( = -5673030240 \)
In simple words: A term is "independent of x" when the variable x disappears, meaning its power becomes zero. Use the general term formula, set the exponent of x to zero, and solve for 'r'. Then substitute 'r' back into the non-x part of the term to find the constant value.

🎯 Exam Tip: Double-check the exponent manipulations for 'x' and ensure that the final constant calculation is accurate. For expressions without the specified variable, note that all terms are independent of it.

Question 4. In the following expansion, find the middle term:
(i) \( {\left(\frac { x }{ 2 }+2y \right)}^{6} \)
(ii) \( {\left(3a-\frac{a^3}{6} \right)}^{9} \)
(iii) \( {\left( x+\frac {1}{x} \right) }^{2n} \)
(iv) \( {\left(3x-\frac{2}{{x}^{2}} \right) }^{15} \)
Answer:
(i) For \( {\left(\frac { x }{ 2 }+2y \right)}^{6} \), the power \( n = 6 \) (which is an even number).
When \( n \) is even, there is one middle term, given by the \( {\left(\frac{n}{2}+1\right)}^{th} \) term.
Here, the middle term is \( {\left(\frac{6}{2}+1\right)}^{th} = (3+1)^{th} = 4^{th} \) term.
Using the general term formula \( T_{r+1} = nC_r A^{n-r} B^r \), for the 4th term, \( r=3 \).
\( A = \frac{x}{2} \), \( B = 2y \), \( n = 6 \).
\( T_4 = T_{3+1} = ^6C_3 {\left(\frac{x}{2}\right)}^{6-3} (2y)^3 \)
\( T_4 = ^6C_3 {\left(\frac{x}{2}\right)}^3 (2y)^3 \)
\( T_4 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{x^3}{2^3} \times 2^3 y^3 \)
\( T_4 = 20 \times \frac{x^3}{8} \times 8 y^3 \)
\( T_4 = 20 x^3 y^3 \)

(ii) For \( {\left(3a-\frac{a^3}{6} \right)}^{9} \), the power \( n = 9 \) (which is an odd number).
When \( n \) is odd, there are two middle terms, given by the \( {\left(\frac{n+1}{2}\right)}^{th} \) term and the \( {\left(\frac{n+1}{2}+1\right)}^{th} \) term.
Here, the middle terms are the \( {\left(\frac{9+1}{2}\right)}^{th} = 5^{th} \) term and the \( {\left(\frac{9+1}{2}+1\right)}^{th} = (5+1)^{th} = 6^{th} \) term.
For the 5th term, \( r=4 \). \( A = 3a \), \( B = -\frac{a^3}{6} \), \( n = 9 \).
\( T_5 = T_{4+1} = ^9C_4 (3a)^{9-4} {\left(-\frac{a^3}{6}\right)}^4 \)
\( T_5 = ^9C_4 (3a)^5 {\left(-\frac{a^3}{6}\right)}^4 \)
\( T_5 = 126 \times 3^5 a^5 \times \frac{a^{12}}{6^4} \)
\( T_5 = 126 \times 243 a^5 \times \frac{a^{12}}{1296} \)
\( T_5 = \frac{30618 a^{17}}{1296} = \frac{189}{8} a^{17} \)
For the 6th term, \( r=5 \). \( A = 3a \), \( B = -\frac{a^3}{6} \), \( n = 9 \).
\( T_6 = T_{5+1} = ^9C_5 (3a)^{9-5} {\left(-\frac{a^3}{6}\right)}^5 \)
\( T_6 = ^9C_5 (3a)^4 {\left(-\frac{a^3}{6}\right)}^5 \)
\( T_6 = 126 \times 3^4 a^4 \times \left(-\frac{a^{15}}{6^5}\right) \)
\( T_6 = 126 \times 81 a^4 \times \left(-\frac{a^{15}}{7776}\right) \)
\( T_6 = -\frac{10206 a^{19}}{7776} = -\frac{21}{16} a^{19} \)
Hence, the two middle terms are \( \frac{189}{8} a^{17} \) and \( -\frac{21}{16} a^{19} \).

(iii) For \( {\left( x+\frac {1}{x} \right) }^{2n} \), the power is \( 2n \) (which is an even number).
When the power is even, there is one middle term, given by the \( {\left(\frac{2n}{2}+1\right)}^{th} = (n+1)^{th} \) term.
For the \( (n+1)^{th} \) term, \( r=n \). \( A = x \), \( B = \frac{1}{x} = x^{-1} \), power \( N=2n \).
\( T_{n+1} = ^{2n}C_n (x)^{2n-n} (x^{-1})^n \)
\( T_{n+1} = ^{2n}C_n x^n x^{-n} \)
\( T_{n+1} = ^{2n}C_n x^{n-n} \)
\( T_{n+1} = ^{2n}C_n x^0 \)
\( T_{n+1} = ^{2n}C_n = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2} \)

(iv) For \( {\left(3x-\frac{2}{{x}^{2}} \right) }^{15} \), the power \( n = 15 \) (which is an odd number).
When the power is odd, there are two middle terms, given by the \( {\left(\frac{15+1}{2}\right)}^{th} = 8^{th} \) term and the \( {\left(\frac{15+1}{2}+1\right)}^{th} = 9^{th} \) term.
For the 8th term, \( r=7 \). \( A = 3x \), \( B = -\frac{2}{x^2} = -2x^{-2} \), \( n=15 \).
\( T_8 = T_{7+1} = ^{15}C_7 (3x)^{15-7} (-2x^{-2})^7 \)
\( T_8 = ^{15}C_7 (3x)^8 (-2x^{-2})^7 \)
\( T_8 = ^{15}C_7 3^8 x^8 (-2)^7 x^{-14} \)
\( T_8 = ^{15}C_7 3^8 (-2)^7 x^{8-14} \)
\( T_8 = 6435 \times 6561 \times (-128) x^{-6} \)
\( T_8 = -538686720 x^{-6} \)
For the 9th term, \( r=8 \). \( A = 3x \), \( B = -\frac{2}{x^2} = -2x^{-2} \), \( n=15 \).
\( T_9 = T_{8+1} = ^{15}C_8 (3x)^{15-8} (-2x^{-2})^8 \)
\( T_9 = ^{15}C_8 (3x)^7 (-2x^{-2})^8 \)
\( T_9 = ^{15}C_8 3^7 x^7 (-2)^8 x^{-16} \)
\( T_9 = ^{15}C_8 3^7 (-2)^8 x^{7-16} \)
\( T_9 = 6435 \times 2187 \times 256 x^{-9} \)
\( T_9 = 3609800640 x^{-9} \)
Hence, the two middle terms are \( -538686720 x^{-6} \) and \( 3609800640 x^{-9} \).
In simple words: The middle term(s) depend on whether the power 'n' is even or odd. If 'n' is even, there is one middle term at position \( (\frac{n}{2}+1) \). If 'n' is odd, there are two middle terms at positions \( (\frac{n+1}{2}) \) and \( (\frac{n+1}{2}+1) \). Once you know 'r', use the general term formula to find the actual term.

🎯 Exam Tip: Always first check if the power 'n' is even or odd to determine the number and position of the middle term(s). Correct calculation of 'r' is crucial before applying the term formula.

Question 5. Prove that if n is even then in expansion of \( (1 + x)^n \), coefficient of middle term will be \( \frac { 1.3.5.....(n-1)}{2.4.6.....n}{2}^{n} \). If n is odd, then coefficient of both the middle term will be \( \frac {1.3.5.....n }{ 2.4.6.....(n+1)}{2}^{n} \).
Answer:
Let's consider the expansion \( (1+x)^N \).
**Case 1: N is even.**
If \( N \) is even, the middle term is \( {\left(\frac{N}{2}+1\right)}^{th} \) term. So \( r = \frac{N}{2} \).
The coefficient of this term is \( ^NC_{N/2} \).
For \( (1+x)^n \) where n is even, we replace \( N \) with \( n \).
The coefficient of the middle term is \( ^nC_{n/2} \).
\( ^nC_{n/2} = \frac{n!}{(n/2)!(n/2)!} = \frac{n!}{((n/2)!)^2} \)
We can write \( n! \) as \( n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1 \).
And \( (n/2)! = (n/2) \times (n/2 - 1) \times ... \times 1 \).
So, \( ^nC_{n/2} = \frac{[1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-1)] \times [2 \cdot 4 \cdot 6 \cdot \ldots \cdot n]}{((n/2)!)^2} \)
\( ^nC_{n/2} = \frac{[1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-1)] \times [2^{n/2} \times (1 \cdot 2 \cdot 3 \cdot \ldots \cdot n/2)]}{((n/2)!)^2} \)
\( ^nC_{n/2} = \frac{[1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-1)] \times 2^{n/2} \times (n/2)!}{((n/2)!)^2} \)
\( ^nC_{n/2} = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-1)}{ (n/2)! } \times 2^{n/2} \)
This can also be written as \( \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-1)}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot n} 2^n \).
So, the coefficient is \( \frac { 1.3.5.....(n-1)}{2.4.6.....n}{2}^{n} \). Hence Proved for even n.

**Case 2: N is odd.**
If \( N \) is odd, there are two middle terms: \( {\left(\frac{N+1}{2}\right)}^{th} \) term and \( {\left(\frac{N+1}{2}+1\right)}^{th} \) term.
For \( (1+x)^n \) where n is odd, we replace \( N \) with \( n \).
The two middle terms are the \( {\left(\frac{n+1}{2}\right)}^{th} \) and \( {\left(\frac{n+1}{2}+1\right)}^{th} \) terms.
The coefficients for these terms are \( ^nC_{(n-1)/2} \) and \( ^nC_{(n+1)/2} \).
Since \( ^nC_r = ^nC_{n-r} \), we have \( ^nC_{(n-1)/2} = ^nC_{n - (n-1)/2} = ^nC_{(n+1)/2} \).
So, both middle terms have the same coefficient: \( ^nC_{(n+1)/2} \).
Let \( k = (n+1)/2 \). Then the coefficient is \( ^nC_k = \frac{n!}{k!(n-k)!} = \frac{n!}{((n+1)/2)!((n-1)/2)!} \).
\( ^nC_{(n+1)/2} = \frac{n \times (n-1) \times \ldots \times 1}{((n+1)/2)! ((n-1)/2)!} \)
\( ^nC_{(n+1)/2} = \frac{[1 \cdot 3 \cdot 5 \cdot \ldots \cdot n] \times [2 \cdot 4 \cdot 6 \cdot \ldots \cdot (n-1)]}{((n+1)/2)! ((n-1)/2)!} \)
\( ^nC_{(n+1)/2} = \frac{[1 \cdot 3 \cdot 5 \cdot \ldots \cdot n] \times [2^{(n-1)/2} \times (1 \cdot 2 \cdot \ldots \cdot (n-1)/2)]}{((n+1)/2)! ((n-1)/2)!} \)
\( ^nC_{(n+1)/2} = \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot n}{((n+1)/2)!} \times 2^{(n-1)/2} \)
This can also be written as \( \frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot n}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (n+1)} 2^n \).
So, the coefficient is \( \frac {1.3.5.....n }{ 2.4.6.....(n+1)}{2}^{n} \). Hence Proved for odd n.
In simple words: The coefficient of the middle term(s) in \( (1+x)^n \) changes depending on whether 'n' is an even or odd number. When 'n' is even, there's one middle term, and its coefficient follows a pattern involving products of odd and even numbers. When 'n' is odd, there are two middle terms, and both have the same coefficient, following a similar pattern.

🎯 Exam Tip: Understand the difference in calculating middle terms for even vs. odd powers 'n'. The factorials in the denominator relate to the product of even numbers, while the numerator contains the product of odd numbers. Be careful with the exponents of 2.

Question 6. If in the expansion of \( {\left(ax+\frac { 1 }{ bx } \right) }^{11} \) coefficient of \( x^7 \) and \( x^{-7} \) are equal then prove that \( ab – 1 = 0 \).
Answer:
The general term \( T_{r+1} \) in the expansion of \( {\left(ax+\frac { 1 }{ bx } \right) }^{11} \) is:
Here, \( A = ax \), \( B = \frac{1}{bx} = b^{-1}x^{-1} \), \( n = 11 \).
\( T_{r+1} = ^{11}C_r (ax)^{11-r} (b^{-1}x^{-1})^r \)
\( T_{r+1} = ^{11}C_r a^{11-r} x^{11-r} b^{-r} x^{-r} \)
\( T_{r+1} = ^{11}C_r a^{11-r} b^{-r} x^{11-r-r} \)
\( T_{r+1} = ^{11}C_r a^{11-r} b^{-r} x^{11-2r} \)

For the coefficient of \( x^7 \):
The power of \( x \) should be \( 7 \).
So, \( 11 - 2r = 7 \)
\( -2r = 7 - 11 \)
\( -2r = -4 \)
\( r = 2 \)
The coefficient of \( x^7 \) is \( ^{11}C_2 a^{11-2} b^{-2} = ^{11}C_2 a^9 b^{-2} = \frac{55a^9}{b^2} \).

For the coefficient of \( x^{-7} \):
The power of \( x \) should be \( -7 \).
So, \( 11 - 2r = -7 \)
\( -2r = -7 - 11 \)
\( -2r = -18 \)
\( r = 9 \)
The coefficient of \( x^{-7} \) is \( ^{11}C_9 a^{11-9} b^{-9} = ^{11}C_9 a^2 b^{-9} = \frac{55a^2}{b^9} \).

According to the question, the coefficients of \( x^7 \) and \( x^{-7} \) are equal:
\( \frac{55a^9}{b^2} = \frac{55a^2}{b^9} \)
Divide both sides by 55:
\( \frac{a^9}{b^2} = \frac{a^2}{b^9} \)
Multiply both sides by \( b^9 \):
\( \frac{a^9 b^9}{b^2} = a^2 \)
\( a^9 b^7 = a^2 \)
Divide both sides by \( a^2 \) (assuming \( a \neq 0 \)):
\( \frac{a^9 b^7}{a^2} = 1 \)
\( a^7 b^7 = 1 \)
\( (ab)^7 = 1 \)
Taking the 7th root of both sides:
\( ab = 1 \)
Subtract 1 from both sides:
\( ab - 1 = 0 \)
Hence Proved.
In simple words: To prove the given statement, first find the general term of the expansion. Then, find the value of 'r' for both \( x^7 \) and \( x^{-7} \) and write down their coefficients. Set these two coefficients equal to each other, and simplify the equation. This will lead to the desired proof.

🎯 Exam Tip: Remember that \( ^nC_r = ^nC_{n-r} \). This property can sometimes simplify calculations. Pay close attention to exponent rules when manipulating terms like \( a^{11-r} \) and \( b^{-r} \).

Question 7. In the expansion of \( (1 + y)^n \) if coefficients of 5th, 6th and 7th terms are in A.P., then find the value of n.
Answer:
For the expansion of \( (1 + y)^n \), the general term is \( T_{r+1} = ^nC_r (1)^{n-r} y^r = ^nC_r y^r \).
So, the coefficient of the \( (r+1)^{th} \) term is \( ^nC_r \).
Coefficient of the 5th term \( (T_5) \) is \( ^nC_4 \).
Coefficient of the 6th term \( (T_6) \) is \( ^nC_5 \).
Coefficient of the 7th term \( (T_7) \) is \( ^nC_6 \).
Given that these coefficients are in Arithmetic Progression (A.P.), it means:
\( 2 \times (\text{middle term}) = (\text{first term}) + (\text{last term}) \)
So, \( 2 \times ^nC_5 = ^nC_4 + ^nC_6 \)
Now, we expand the combinatorial terms:
\( 2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \)
Divide all terms by \( n! \) (assuming \( n! \neq 0 \)):
\( \frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \)
Rewrite the factorials to find a common denominator:
\( \frac{2}{5 \times 4!(n-5)!} = \frac{1}{4!(n-4)(n-5)!} + \frac{1}{6 \times 5 \times 4!(n-6)!} \)
Multiply the entire equation by \( 4!(n-6)! \):
\( \frac{2}{5(n-5)(n-6)!} = \frac{1}{(n-4)(n-5)(n-6)!} + \frac{1}{30(n-6)!} \)
Let's simplify:
\( \frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30} \)
Multiply the entire equation by \( 30(n-4)(n-5) \) to clear denominators:
\( 2 \times 6(n-4) = 30 + (n-4)(n-5) \)
\( 12(n-4) = 30 + n^2 - 5n - 4n + 20 \)
\( 12n - 48 = 30 + n^2 - 9n + 20 \)
Rearrange into a quadratic equation:
\( n^2 - 9n - 12n + 30 + 20 + 48 = 0 \)
\( n^2 - 21n + 98 = 0 \)
Factor the quadratic equation:
We need two numbers that multiply to 98 and add to -21. These are -7 and -14.
\( (n-7)(n-14) = 0 \)
So, \( n-7=0 \implies n=7 \)
Or, \( n-14=0 \implies n=14 \)
Both values are valid since \( n \ge 6 \) (because we have \( ^nC_6 \) as a term).
Therefore, the value of n can be 7 or 14.
In simple words: When coefficients of consecutive terms in a binomial expansion are in an arithmetic progression, the middle coefficient is the average of the coefficients before and after it. By setting up this relationship using combinatorial formulas and simplifying, you can solve for 'n'.

🎯 Exam Tip: Remember the property of A.P.: If a, b, c are in A.P., then \( 2b = a+c \). Be careful when simplifying factorials; expand the larger factorial down to the smaller one (e.g., \( (n-4)! = (n-4)(n-5)! \)) to cancel terms efficiently.

Question 8. In Binomial expansion \( (x – a)^n \) second, third and fourth terms are 250, 720 and 1080 respectively. Find x, a and n.
Answer:
The general term \( T_{r+1} \) in the expansion of \( (x – a)^n \) is \( T_{r+1} = ^nC_r (x)^{n-r} (-a)^r \).
Second term \( (T_2) \): Here \( r=1 \).
\( T_2 = ^nC_1 (x)^{n-1} (-a)^1 = -^nC_1 x^{n-1} a = -nx^{n-1}a = 250 \) ... (1)
Third term \( (T_3) \): Here \( r=2 \).
\( T_3 = ^nC_2 (x)^{n-2} (-a)^2 = ^nC_2 x^{n-2} a^2 = 720 \) ... (2)
Fourth term \( (T_4) \): Here \( r=3 \).
\( T_4 = ^nC_3 (x)^{n-3} (-a)^3 = -^nC_3 x^{n-3} a^3 = 1080 \) ... (3)

Divide equation (2) by equation (1):
\( \frac{T_3}{T_2} = \frac{^nC_2 x^{n-2} a^2}{-^nC_1 x^{n-1} a} = \frac{720}{250} \)
\( \frac{\frac{n(n-1)}{2} x^{n-2} a^2}{-n x^{n-1} a} = \frac{72}{25} \)
\( -\frac{(n-1)a}{2x} = \frac{72}{25} \)
\( \frac{(n-1)a}{x} = -\frac{144}{25} \) ... (A)

Divide equation (3) by equation (2):
\( \frac{T_4}{T_3} = \frac{-^nC_3 x^{n-3} a^3}{^nC_2 x^{n-2} a^2} = \frac{1080}{720} \)
\( \frac{-\frac{n(n-1)(n-2)}{6} x^{n-3} a^3}{\frac{n(n-1)}{2} x^{n-2} a^2} = \frac{108}{72} \)
\( -\frac{(n-2)a}{3x} = \frac{3}{2} \)
\( \frac{(n-2)a}{x} = -\frac{9}{2} \) ... (B)

Now, divide equation (A) by equation (B):
\( \frac{\frac{(n-1)a}{x}}{\frac{(n-2)a}{x}} = \frac{-\frac{144}{25}}{-\frac{9}{2}} \)
\( \frac{n-1}{n-2} = \frac{144}{25} \times \frac{2}{9} \)
\( \frac{n-1}{n-2} = \frac{16 \times 2}{25} \)
\( \frac{n-1}{n-2} = \frac{32}{25} \)
Cross-multiply:
\( 25(n-1) = 32(n-2) \)
\( 25n - 25 = 32n - 64 \)
\( 64 - 25 = 32n - 25n \)
\( 39 = 7n \)
\( n = \frac{39}{7} \)
This result for 'n' is not an integer. Let's recheck the problem or calculations. The problem states \( (x-a)^n \). If the terms are positive (250, 720, 1080), the original signs must be correct. Equation (1) \( -nx^{n-1}a = 250 \) means one of n, x, a must be negative or an odd number of them. Equation (2) \( \frac{n(n-1)}{2} x^{n-2} a^2 = 720 \) - always positive if x and a are real. Equation (3) \( -\frac{n(n-1)(n-2)}{6} x^{n-3} a^3 = 1080 \). This should be negative if n,x,a are positive. There is a sign inconsistency here if x,a are real and positive. The 2nd term should be negative, the 3rd positive, the 4th negative. Given the terms are 250, 720, 1080, let's assume the magnitudes are given, and the signs were just omitted or inverted in the source for T2 and T4, or the original expansion was \( (x+a)^n \). If we follow the source's interpretation of signs (i.e., T2, T3, T4 are all positive values as given), this implies \( -nx^{n-1}a \) must be 250, which means one of x or a (or both) must be such that the expression becomes positive. Let's assume the base expansion is \( (x+a)^n \). \( T_{r+1} = ^nC_r x^{n-r} a^r \). Then: \( T_2 = ^nC_1 x^{n-1} a = nx^{n-1}a = 250 \) ... (1') \( T_3 = ^nC_2 x^{n-2} a^2 = 720 \) ... (2') \( T_4 = ^nC_3 x^{n-3} a^3 = 1080 \) ... (3') Divide (2') by (1'): \( \frac{T_3}{T_2} = \frac{^nC_2 x^{n-2} a^2}{^nC_1 x^{n-1} a} = \frac{720}{250} \) \( \frac{n(n-1)/2}{n} \frac{a}{x} = \frac{72}{25} \) \( \frac{n-1}{2} \frac{a}{x} = \frac{72}{25} \) \( \frac{(n-1)a}{x} = \frac{144}{25} \) ... (A') Divide (3') by (2'): \( \frac{T_4}{T_3} = \frac{^nC_3 x^{n-3} a^3}{^nC_2 x^{n-2} a^2} = \frac{1080}{720} \) \( \frac{n(n-1)(n-2)/6}{n(n-1)/2} \frac{a}{x} = \frac{3}{2} \) \( \frac{n-2}{3} \frac{a}{x} = \frac{3}{2} \) \( \frac{(n-2)a}{x} = \frac{9}{2} \) ... (B') Divide (A') by (B'): \( \frac{\frac{(n-1)a}{x}}{\frac{(n-2)a}{x}} = \frac{144/25}{9/2} \) \( \frac{n-1}{n-2} = \frac{144}{25} \times \frac{2}{9} = \frac{16 \times 2}{25} = \frac{32}{25} \) \( 25(n-1) = 32(n-2) \) \( 25n - 25 = 32n - 64 \)
\( 7n = 39 \)
\( n = \frac{39}{7} \)
The value of 'n' still comes out as \( \frac{39}{7} \), which is not a natural number, but 'n' must be a natural number for binomial expansion. This suggests there might be an error in the problem statement or the given term values. Let's re-examine the source material's image for any discrepancies. The solution presented in the OCR snippet on page 15 has: \( \frac{a}{x} = \frac{6}{n-1} \) from (4) and \( \frac{a}{x} = \frac{9}{2(n-2)} \) from (5). Setting these equal: \( \frac{6}{n-1} = \frac{9}{2(n-2)} \) \( 12(n-2) = 9(n-1) \) \( 12n - 24 = 9n - 9 \) \( 3n = 15 \) \( n = 5 \)
Now, let's find a and x using n=5.
From \( \frac{a}{x} = \frac{6}{n-1} \):
\( \frac{a}{x} = \frac{6}{5-1} = \frac{6}{4} = \frac{3}{2} \)
So, \( 2a = 3x \).
Substitute \( n=5 \) into (1') \( nx^{n-1}a = 250 \):
\( 5x^{5-1}a = 250 \)
\( 5x^4a = 250 \)
\( x^4a = 50 \)
We have \( a = \frac{3}{2}x \). Substitute this into \( x^4a = 50 \):
\( x^4 \left(\frac{3}{2}x\right) = 50 \)
\( \frac{3}{2} x^5 = 50 \)
\( x^5 = \frac{100}{3} \). This would give \( x = \sqrt[5]{\frac{100}{3}} \), which is not an integer.
The OCR solution on page 15 then says "From \( 5x^4a = 240 \) and from (4) \( \frac{a}{x} = \frac{3}{2} \)". It uses 240 instead of 250. If \( 5x^4a = 240 \), then \( x^4a = 48 \).
\( x^4 \left(\frac{3}{2}x\right) = 48 \)
\( \frac{3}{2} x^5 = 48 \)
\( 3x^5 = 96 \)
\( x^5 = 32 \)
\( x = 2 \)
If \( x=2 \), then \( a = \frac{3}{2}x = \frac{3}{2}(2) = 3 \).
So, \( n=5, x=2, a=3 \). Let's check these values with the original equations assuming \( (x+a)^n \):
\( T_2 = nx^{n-1}a = 5(2)^{5-1}(3) = 5(2)^4(3) = 5(16)(3) = 240 \). (Matches if T2 was 240, not 250)
\( T_3 = ^nC_2 x^{n-2} a^2 = ^5C_2 (2)^{5-2}(3)^2 = 10(2)^3(9) = 10(8)(9) = 720 \). (Matches)
\( T_4 = ^nC_3 x^{n-3} a^3 = ^5C_3 (2)^{5-3}(3)^3 = 10(2)^2(27) = 10(4)(27) = 1080 \). (Matches)
So, it seems the original \( T_2 \) value was a typo and should have been 240, and the expansion is \( (x+a)^n \). I will proceed with \( n=5, x=2, a=3 \).
In simple words: Write down the algebraic expressions for the second, third, and fourth terms using the general binomial term formula. Set these equal to the given numerical values. Divide the equations by each other to eliminate variables and find 'n'. Then, use the value of 'n' to find 'x' and 'a'. Be careful with signs in the binomial expansion \( (x-a)^n \).

🎯 Exam Tip: When given multiple terms, dividing successive terms is a common strategy to eliminate \( ^nC_r \) and \( x \) or \( a \) terms, making it easier to solve for 'n'. Always double-check your sign conventions and if the terms are positive, consider if the original expansion might be \( (x+a)^n \).

Question 9. If in the expansion of (1+ a) coefficient of three consecutive terms are in ratio 1 : 7 ; 42, then find the value of n.
Answer: Let the three consecutive terms in the expansion of \( (1+a)^n \) be \( T_{r-1} \), \( T_r \), and \( T_{r+1} \). Their coefficients are \( ^nC_{r-2} \), \( ^nC_{r-1} \), and \( ^nC_r \) respectively.
The given ratio of coefficients is \( ^nC_{r-2} : ^nC_{r-1} : ^nC_r = 1 : 7 : 42 \).
First ratio: \( \frac{^nC_{r-2}}{^nC_{r-1}} = \frac{1}{7} \)
Using the identity \( \frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k} \), we get:
\( \frac{r-1}{n-(r-1)+1} = \frac{1}{7} \)
\( \frac{r-1}{n-r+2} = \frac{1}{7} \)
\( 7(r-1) = n-r+2 \)
\( 7r-7 = n-r+2 \)
\( n-8r+9 = 0 \) ...(1)

Second ratio: \( \frac{^nC_{r-1}}{^nC_r} = \frac{7}{42} = \frac{1}{6} \)
Using the identity \( \frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k} \), we get:
\( \frac{r}{n-r+1} = \frac{1}{6} \)
\( 6r = n-r+1 \)
\( n-7r+1 = 0 \) ...(2)

Now, we solve equations (1) and (2):
Subtract equation (2) from equation (1):
\( (n-8r+9) - (n-7r+1) = 0 \)
\( n-8r+9-n+7r-1 = 0 \)
\( -r+8 = 0 \)
\( r = 8 \)

Substitute \( r=8 \) into equation (2):
\( n-7(8)+1 = 0 \)
\( n-56+1 = 0 \)
\( n-55 = 0 \)
\( n = 55 \)
In simple words: We used the given ratio of the coefficients for three terms in a binomial expansion. By setting up two equations based on these ratios and solving them, we found the value of 'n'.

🎯 Exam Tip: Remember the formula for the ratio of consecutive binomial coefficients: \( \frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} \). This is key for solving such problems efficiently.

Question 10. Find the positive value of m for which in the expansion of (1 + x)m coefficient x² is 6.
Answer: We know that the binomial expansion of \( (1+x)^m \) is given by:
\( (1+x)^m = ^mC_0 + ^mC_1x + ^mC_2x^2 + ^mC_3x^3 + ... \)
The coefficient of \( x^2 \) in this expansion is \( ^mC_2 \).
According to the question, the coefficient of \( x^2 \) is 6.
So, \( ^mC_2 = 6 \)
We use the formula for combinations: \( ^mC_r = \frac{m!}{r!(m-r)!} \)
For \( ^mC_2 \), we have:
\( \frac{m(m-1)(m-2)!}{2!(m-2)!} = 6 \)
\( \frac{m(m-1)}{2 \times 1} = 6 \)
\( m(m-1) = 12 \)
Now, expand and rearrange the equation to form a quadratic equation:
\( m^2 - m = 12 \)
\( m^2 - m - 12 = 0 \)
We need to factorize this quadratic equation. We look for two numbers that multiply to -12 and add up to -1. These numbers are -4 and 3.
\( m^2 - 4m + 3m - 12 = 0 \)
\( m(m-4) + 3(m-4) = 0 \)
\( (m-4)(m+3) = 0 \)
This gives two possible values for m:
\( m-4 = 0 \implies m = 4 \)
\( m+3 = 0 \implies m = -3 \)
The question asks for the *positive* value of m. Therefore, we choose \( m=4 \).
In simple words: We used the formula for the coefficient of \(x^2\) in the binomial expansion, set it equal to 6, and then solved the resulting equation to find the value of 'm'. Since we needed a positive answer, we picked m=4.

🎯 Exam Tip: Always remember to consider the conditions given in the question (like "positive value") when selecting the final answer from multiple solutions.