Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.6

Samacheer Kalvi 11th Business Maths Algebra Ex 2.6 Text Book Back Questions And Answers

Question 1. Expand the following by using binomial theorem:
(i) \( (2a - 3b)^4 \)
(ii) \( \left(x+\frac{1}{y}\right)^{7} \)
(iii) \( \left(x+\frac{1}{x^{2}}\right)^{6} \)
Answer:
(i) To expand \( (2a - 3b)^4 \), we use the binomial theorem. The theorem helps us expand expressions of the form \( (x + y)^n \). For this problem, \( x \) is \( 2a \), \( y \) is \( -3b \), and \( n \) is \( 4 \). We apply the formula, calculating each term. The coefficients are found using combinations \( (nC_r) \). This makes it easy to handle higher powers.
\( (2a-3b)^4 = (4C_0)(2a)^4 (-3b)^0 + (4C_1)(2a)^3 (-3b)^1 + (4C_2)(2a)^2 (-3b)^2 + (4C_3)(2a)^1 (-3b)^3 + (4C_4)(2a)^0 (-3b)^4 \)
\( = 1 \cdot (16a^4) \cdot 1 + 4 \cdot (8a^3) \cdot (-3b) + 6 \cdot (4a^2) \cdot (9b^2) + 4 \cdot (2a) \cdot (-27b^3) + 1 \cdot 1 \cdot (81b^4) \)
\( = 16a^4 - 96a^3b + 216a^2b^2 - 216ab^3 + 81b^4 \)
(ii) Here, \( x \) is \( x \), \( a \) is \( \frac{1}{y} \), and \( n \) is \( 7 \). We apply the binomial expansion formula. Each term includes a combination coefficient and powers of \( x \) and \( \frac{1}{y} \). We then simplify each term by multiplying the numbers and combining the powers.
\( \left(x+\frac{1}{y}\right)^{7} = (7C_0)x^7 \left(\frac{1}{y}\right)^0 + (7C_1)x^6 \left(\frac{1}{y}\right)^1 + (7C_2)x^5 \left(\frac{1}{y}\right)^2 + (7C_3)x^4 \left(\frac{1}{y}\right)^3 + (7C_4)x^3 \left(\frac{1}{y}\right)^4 + (7C_5)x^2 \left(\frac{1}{y}\right)^5 + (7C_6)x^1 \left(\frac{1}{y}\right)^6 + (7C_7)x^0 \left(\frac{1}{y}\right)^7 \)
\( = x^7 + 7x^6 \left(\frac{1}{y}\right) + 21x^5 \left(\frac{1}{y^2}\right) + 35x^4 \left(\frac{1}{y^3}\right) + 35x^3 \left(\frac{1}{y^4}\right) + 21x^2 \left(\frac{1}{y^5}\right) + 7x \left(\frac{1}{y^6}\right) + 1 \left(\frac{1}{y^7}\right) \)
\( = x^7 + \frac{7x^6}{y} + \frac{21x^5}{y^2} + \frac{35x^4}{y^3} + \frac{35x^3}{y^4} + \frac{21x^2}{y^5} + \frac{7x}{y^6} + \frac{1}{y^7} \)
(iii) For this expansion, \( x \) is \( x \), \( a \) is \( \frac{1}{x^2} \), and \( n \) is \( 6 \). We use the binomial theorem again. After applying the formula and simplifying, we combine the powers of \( x \). Remember to subtract the exponents when dividing.
\( \left(x+\frac{1}{x^2}\right)^{6} = (6C_0)x^6 \left(\frac{1}{x^2}\right)^0 + (6C_1)x^5 \left(\frac{1}{x^2}\right)^1 + (6C_2)x^4 \left(\frac{1}{x^2}\right)^2 + (6C_3)x^3 \left(\frac{1}{x^2}\right)^3 + (6C_4)x^2 \left(\frac{1}{x^2}\right)^4 + (6C_5)x^1 \left(\frac{1}{x^2}\right)^5 + (6C_6)x^0 \left(\frac{1}{x^2}\right)^6 \)
\( = x^6 + 6x^5 \left(\frac{1}{x^2}\right) + 15x^4 \left(\frac{1}{x^4}\right) + 20x^3 \left(\frac{1}{x^6}\right) + 15x^2 \left(\frac{1}{x^8}\right) + 6x \left(\frac{1}{x^{10}}\right) + 1 \left(\frac{1}{x^{12}}\right) \)
\( = x^6 + 6x^3 + 15 + \frac{20}{x^3} + \frac{15}{x^6} + \frac{6}{x^9} + \frac{1}{x^{12}} \)
In simple words: The binomial theorem helps us expand expressions with powers. We use combinations to find the numbers for each term, then combine the letters and their powers. Be careful with minus signs and fractions.

🎯 Exam Tip: Always write down the general term formula for binomial expansion and clearly identify the values of `x`, `a`, and `n` for each sub-question before proceeding with calculations.

Question 2. Evaluate the following using binomial theorem:
(i) \( (101)^4 \)
(ii) \( (999)^5 \)
Answer:
(i) To evaluate \( (101)^4 \), we write it as \( (100 + 1)^4 \). Then, we use the binomial theorem where \( x \) is \( 100 \), \( a \) is \( 1 \), and \( n \) is \( 4 \). We expand each term, calculate the powers and combinations, and then add all the results together. This method makes large number calculations easier.
\( (101)^4 = (100 + 1)^4 \)
\( = (4C_0)(100)^4 (1)^0 + (4C_1)(100)^3 (1)^1 + (4C_2)(100)^2 (1)^2 + (4C_3)(100)^1 (1)^3 + (4C_4)(100)^0 (1)^4 \)
\( = 1 \cdot (100000000) + 4 \cdot (1000000) \cdot 1 + 6 \cdot (10000) \cdot 1 + 4 \cdot (100) \cdot 1 + 1 \cdot 1 \cdot 1 \)
\( = 100000000 + 4000000 + 60000 + 400 + 1 \)
\( = 104060401 \)
(ii) To evaluate \( (999)^5 \), we rewrite it as \( (1000 - 1)^5 \). Here, \( x \) is \( 1000 \), \( a \) is \( -1 \), and \( n \) is \( 5 \). We apply the binomial theorem, paying close attention to the alternating signs due to \( (-1) \) terms. We calculate each expanded part and then combine them to get the final answer. This helps simplify calculations involving numbers close to powers of 10.
\( (999)^5 = (1000 - 1)^5 \)
\( = (5C_0)(1000)^5 (-1)^0 + (5C_1)(1000)^4 (-1)^1 + (5C_2)(1000)^3 (-1)^2 + (5C_3)(1000)^2 (-1)^3 + (5C_4)(1000)^1 (-1)^4 + (5C_5)(1000)^0 (-1)^5 \)
\( = 1 \cdot (1000)^5 - 5 \cdot (1000)^4 \cdot 1 + 10 \cdot (1000)^3 \cdot 1 - 10 \cdot (1000)^2 \cdot 1 + 5 \cdot (1000)^1 \cdot 1 - 1 \cdot 1 \cdot 1 \)
\( = 1000000000000000 - 5000000000000 + 10000000000 - 10000000 + 5000 - 1 \)
\( = 995009990004999 \)
In simple words: When a number is close to a power of ten, like 101 or 999, you can write it as (100+1) or (1000-1). Then use the binomial formula to calculate it. It makes big multiplications easier to solve.

🎯 Exam Tip: Always express numbers like 101 or 999 as a sum or difference with powers of 10 (e.g., 100+1 or 1000-1). This simplifies the calculation significantly by using the binomial theorem effectively.

Question 3. Find the 5th term in the expansion of \( (x – 2y)^{13} \).
Answer: To find a specific term in a binomial expansion, we use the general term formula: \( t_{r+1} = nC_r x^{n-r} a^r \). Here, for the 5th term, \( r \) will be 4. We substitute \( n=13 \), \( x=x \), and \( a=-2y \) into the formula. Then, we calculate the combination and simplify the powers to get the final expression for the term. This formula is very helpful for finding any specific term without expanding the whole expression.
Given expansion is \( (x - 2y)^{13} \).
Comparing with \( (X + A)^N \), we have \( N = 13 \), \( X = x \), \( A = -2y \).
To find the 5th term, we use \( t_{r+1} \) where \( r+1 = 5 \), so \( r = 4 \).
The general term formula is \( t_{r+1} = N C_r X^{N-r} A^r \).
Substitute the values:
\( t_{4+1} = t_5 = (13C_4) (x)^{13-4} (-2y)^4 \)
\( t_5 = (13C_4) (x)^9 (-2)^4 (y)^4 \)
\( 13C_4 = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 13 \times 11 \times 5 = 715 \)
\( (-2)^4 = 16 \)
\( t_5 = 715 \cdot x^9 \cdot 16 \cdot y^4 \)
\( t_5 = 11440 x^9 y^4 \)
In simple words: To find a specific term in an expansion, like the 5th term, we use a special formula. We put the total power (13), the term number (5, so r=4), and the parts of the expression (x and -2y) into the formula. Then we calculate the numbers and powers.

🎯 Exam Tip: When finding a specific term \( t_{r+1} \), remember that \( r \) is always one less than the term number. Pay close attention to negative signs in the term `a`, as they affect the sign of the final result.

Question 4. Find the middle terms in the expansion of
(i) \( \left(x+\frac{1}{x}\right)^{11} \)
(ii) \( \left(3 x+\frac{x^{2}}{2}\right)^{8} \)
(iii) \( \left(2 x^{2}-\frac{3}{x^{3}}\right)^{10} \)
Answer:
(i) When the power \( n \) is odd, there are two middle terms in the binomial expansion. These are found using the formulas \( t_{(n+1)/2} \) and \( t_{(n+3)/2} \). For \( n=11 \), the middle terms are the 6th and 7th terms. We apply the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \) for each case, using \( r=5 \) and \( r=6 \) respectively, and then simplify the expressions. This way, we identify the exact middle parts of the expansion.
Given expansion is \( \left(x+\frac{1}{x}\right)^{11} \). Here \( n=11 \) (odd).
The middle terms are \( t_{\frac{11+1}{2}} \) and \( t_{\frac{11+3}{2}} \), which are \( t_6 \) and \( t_7 \).
For \( t_6 \): \( r = 5 \). \( x = x \), \( a = \frac{1}{x} \).
\( t_6 = (11C_5) (x)^{11-5} \left(\frac{1}{x}\right)^5 \)
\( t_6 = (11C_5) x^6 \cdot \frac{1}{x^5} \)
\( t_6 = (11C_5) x^{6-5} \)
\( t_6 = 462x \)
For \( t_7 \): \( r = 6 \). \( x = x \), \( a = \frac{1}{x} \).
\( t_7 = (11C_6) (x)^{11-6} \left(\frac{1}{x}\right)^6 \)
\( t_7 = (11C_6) x^5 \cdot \frac{1}{x^6} \)
\( t_7 = (11C_6) x^{5-6} \)
\( t_7 = 462x^{-1} = \frac{462}{x} \)
(ii) When the power \( n \) is even, there is only one middle term. This term is found using the formula \( t_{n/2 + 1} \). For \( n=8 \), the middle term is the 5th term. We use the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \), setting \( r=4 \), \( x=3x \), and \( a=\frac{x^2}{2} \). After calculating the combination and simplifying the powers of \( x \), we get the final expression for the middle term. This helps in quickly locating the central part of an even-powered expansion.
Given expansion is \( \left(3 x+\frac{x^{2}}{2}\right)^{8} \). Here \( n=8 \) (even).
The only middle term is \( t_{\frac{8}{2}+1} \), which is \( t_5 \).
For \( t_5 \): \( r = 4 \). \( x = 3x \), \( a = \frac{x^2}{2} \).
\( t_5 = (8C_4) (3x)^{8-4} \left(\frac{x^2}{2}\right)^4 \)
\( t_5 = (8C_4) (3x)^4 \left(\frac{x^8}{16}\right) \)
\( t_5 = (8C_4) (81x^4) \left(\frac{x^8}{16}\right) \)
\( t_5 = \frac{70 \times 81}{16} x^{4+8} \)
\( t_5 = \frac{2835}{8} x^{12} \)
(iii) For \( n=10 \), which is an even number, we find the single middle term using \( t_{n/2 + 1} \), which gives us the 6th term. We then apply the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \), with \( r=5 \), \( x=2x^2 \), and \( a=-\frac{3}{x^3} \). We carefully simplify the powers and coefficients. The negative sign in \( a \) leads to an alternating sign in the expansion. The terms are combined to get the simplified middle term. This calculation shows how to manage negative terms and powers of x in the denominator.
Given expansion is \( \left(2 x^{2}-\frac{3}{x^{3}}\right)^{10} \). Here \( n=10 \) (even).
The only middle term is \( t_{\frac{10}{2}+1} \), which is \( t_6 \).
For \( t_6 \): \( r = 5 \). \( x = 2x^2 \), \( a = -\frac{3}{x^3} \).
\( t_6 = (10C_5) (2x^2)^{10-5} \left(-\frac{3}{x^3}\right)^5 \)
\( t_6 = (10C_5) (2x^2)^5 \left(-\frac{3}{x^3}\right)^5 \)
\( t_6 = (10C_5) (2^5 x^{10}) \left(\frac{(-3)^5}{x^{15}}\right) \)
\( t_6 = (10C_5) (32 x^{10}) \left(\frac{-243}{x^{15}}\right) \)
\( t_6 = (10C_5) (32) (-243) x^{10-15} \)
\( t_6 = 252 \times (-7776) x^{-5} \)
\( t_6 = -1959552 x^{-5} = -\frac{1959552}{x^5} \)
In simple words: To find the middle terms, first check if the power is odd or even. If odd, there are two middle terms. If even, there is one middle term. Then use the general term formula for each. Remember to correctly combine the powers of 'x'.

🎯 Exam Tip: Always determine if `n` is odd or even first, as this dictates if there is one or two middle terms. For terms with `x` in the denominator, be careful with negative exponents when combining powers.

Question 5. Find the term independent of x in the expansion of
(i) \( \left(x^{2}-\frac{2}{3 x}\right)^{9} \)
(ii) \( \left(x-\frac{2}{x^{2}}\right)^{15} \)
(iii) \( \left(2 x^{2}+\frac{1}{x}\right)^{12} \)
Answer:
(i) To find the term independent of \( x \) in a binomial expansion, we use the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \). We set \( x = x^2 \), \( a = -\frac{2}{3x} \), and \( n = 9 \). We combine all powers of \( x \) in the general term. For the term to be independent of \( x \), the exponent of \( x \) must be zero. We solve for \( r \) and then substitute this value back into the general term formula to find the specific term. This isolates the constant part of the expansion. After calculation and simplification, we get the numerical value.
Given expansion is \( \left(x^{2}-\frac{2}{3 x}\right)^{9} \).
General term \( t_{r+1} = (9C_r) (x^2)^{9-r} \left(-\frac{2}{3x}\right)^r \)
\( t_{r+1} = (9C_r) x^{2(9-r)} \frac{(-2)^r}{3^r x^r} \)
\( t_{r+1} = (9C_r) \frac{(-2)^r}{3^r} x^{18-2r-r} \)
\( t_{r+1} = (9C_r) \frac{(-2)^r}{3^r} x^{18-3r} \)
For the term independent of \( x \), the power of \( x \) must be zero.
\( 18 - 3r = 0 \)
\( 3r = 18 \)
\( \implies r = 6 \)
Substitute \( r=6 \) into the term:
\( t_{6+1} = t_7 = (9C_6) \frac{(-2)^6}{3^6} \)
\( t_7 = (9C_3) \frac{64}{729} \) (since \( 9C_6 = 9C_{9-6} = 9C_3 \))
\( 9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \)
\( t_7 = 84 \times \frac{64}{729} = \frac{5376}{729} = \frac{1792}{243} \)
(ii) To find the term independent of \( x \) in this expansion, we start with the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \). Here, \( n=15 \), \( x=x \), and \( a=-\frac{2}{x^2} \). We collect all powers of \( x \) together. For the term to be independent of \( x \), the combined exponent of \( x \) must be zero. We solve this equation for \( r \) and then plug that \( r \) value back into the general term formula to find the required term. This method efficiently extracts the constant term from the binomial series. The negative power means the term will be negative.
Given expansion is \( \left(x-\frac{2}{x^{2}}\right)^{15} \).
General term \( t_{r+1} = (15C_r) (x)^{15-r} \left(-\frac{2}{x^2}\right)^r \)
\( t_{r+1} = (15C_r) x^{15-r} \frac{(-2)^r}{(x^2)^r} \)
\( t_{r+1} = (15C_r) (-2)^r x^{15-r-2r} \)
\( t_{r+1} = (15C_r) (-2)^r x^{15-3r} \)
For the term independent of \( x \), the power of \( x \) must be zero.
\( 15 - 3r = 0 \)
\( 3r = 15 \)
\( \implies r = 5 \)
Substitute \( r=5 \) into the term:
\( t_{5+1} = t_6 = (15C_5) (-2)^5 \)
\( 15C_5 = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = 3 \times 7 \times 13 \times 11 = 3003 \)
\( (-2)^5 = -32 \)
\( t_6 = 3003 \times (-32) = -96096 \)
(iii) To find the term independent of \( x \) in this binomial expansion, we begin with the general term formula \( t_{r+1} = nC_r x^{n-r} a^r \). Here, \( n=12 \), \( x=2x^2 \), and \( a=\frac{1}{x} \). We simplify the expression by combining all powers of \( x \). For the term to be independent of \( x \), its exponent must be zero. We solve for \( r \) and then use this value to calculate the specific term. This process helps us find the constant numerical value within the expanded series. The calculation involves binomial coefficients and powers of a constant.
Given expansion is \( \left(2 x^{2}+\frac{1}{x}\right)^{12} \).
General term \( t_{r+1} = (12C_r) (2x^2)^{12-r} \left(\frac{1}{x}\right)^r \)
\( t_{r+1} = (12C_r) 2^{12-r} (x^2)^{12-r} \frac{1}{x^r} \)
\( t_{r+1} = (12C_r) 2^{12-r} x^{2(12-r)} x^{-r} \)
\( t_{r+1} = (12C_r) 2^{12-r} x^{24-2r-r} \)
\( t_{r+1} = (12C_r) 2^{12-r} x^{24-3r} \)
For the term independent of \( x \), the power of \( x \) must be zero.
\( 24 - 3r = 0 \)
\( 3r = 24 \)
\( \implies r = 8 \)
Substitute \( r=8 \) into the term:
\( t_{8+1} = t_9 = (12C_8) 2^{12-8} \)
\( t_9 = (12C_4) 2^4 \) (since \( 12C_8 = 12C_{12-8} = 12C_4 \))
\( 12C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 11 \times 5 \times 9 = 495 \)
\( 2^4 = 16 \)
\( t_9 = 495 \times 16 = 7920 \)
In simple words: A term "independent of x" means it has no 'x' in it, so the power of 'x' is zero. You find the general term, combine all 'x' powers, set that total power to zero to find 'r', then put 'r' back into the term to get the final number.

🎯 Exam Tip: The key to finding the term independent of `x` is to combine all exponents of `x` in the general term and set the resulting expression equal to zero to solve for `r`. Always double-check your exponent arithmetic.

Question 6. Prove that the term independent of x in the expansion of \( \left(x+\frac{1}{x}\right)^{2 n} \) is \( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1) 2^{n}}{n !} \).
Answer: To prove the given statement, we first identify that the expansion \( \left(x+\frac{1}{x}\right)^{2n} \) has an even power (\( 2n \)), so there is exactly one middle term, which is the term independent of \( x \). This term is \( t_{n+1} \), meaning \( r=n \). We apply the general term formula \( t_{r+1} = N C_r X^{N-r} A^r \) with \( N=2n \), \( X=x \), \( A=\frac{1}{x} \), and \( r=n \). This simplifies to \( (2n)C_n \). Next, we expand \( (2n)C_n \) using its factorial definition. We separate the even and odd terms in the numerator and factor out \( 2^n \) from the even terms. After canceling \( n! \), we arrive at the desired expression. This shows the term independent of x is exactly as given.
Given expansion is \( \left(x+\frac{1}{x}\right)^{2n} \).
Here, the power is \( N = 2n \). Since \( N \) is even, there is one middle term.
The middle term is \( t_{\frac{2n}{2}+1} = t_{n+1} \).
For \( t_{n+1} \), we have \( r=n \).
Using the general term formula \( t_{r+1} = N C_r X^{N-r} A^r \):
\( t_{n+1} = (2nC_n) (x)^{2n-n} \left(\frac{1}{x}\right)^n \)
\( t_{n+1} = (2nC_n) x^n \cdot \frac{1}{x^n} \)
\( t_{n+1} = (2nC_n) \)
Now, we expand \( (2nC_n) \):
\( (2nC_n) = \frac{(2n)!}{n! n!} \)
\( = \frac{(2n) \cdot (2n-1) \cdot (2n-2) \cdot \cdot \cdot 4 \cdot 3 \cdot 2 \cdot 1}{n! n!} \)
\( = \frac{[ (2n) \cdot (2n-2) \cdot \cdot \cdot 4 \cdot 2 ] \cdot [ (2n-1) \cdot (2n-3) \cdot \cdot \cdot 3 \cdot 1 ]}{n! n!} \)
\( = \frac{2^n [ n \cdot (n-1) \cdot \cdot \cdot 2 \cdot 1 ] \cdot [ 1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) ]}{n! n!} \)
\( = \frac{2^n \cdot n! \cdot [ 1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) ]}{n! n!} \)
\( = \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) 2^n}{n!} \)
Hence proved.
In simple words: For an expression like \( (x + 1/x)^{2n} \), the term that doesn't have 'x' in it is always the middle term. We use a formula for this term and break it down using factorials to show it matches the given complex fraction.

🎯 Exam Tip: For proof questions, ensure each step is logical and clearly derived. When expanding factorials, grouping even and odd terms can simplify the expression and lead directly to the desired form.

Question 7. Show that the middle term in the expansion of \( (1 + x)^{2n} \) is \( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1) 2^{n}}{n !} x^n \).
Answer: To show the middle term of \( (1+x)^{2n} \), we first recognize that since the power \( 2n \) is even, there is one middle term, which is \( t_{n+1} \). We use the general term formula \( t_{r+1} = N C_r X^{N-r} A^r \) with \( N=2n \), \( X=1 \), \( A=x \), and \( r=n \). This gives us \( (2n)C_n x^n \). We recall from the previous question that \( (2n)C_n \) can be expanded and simplified to \( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) 2^n}{n!} \). Substituting this back, we prove the expression for the middle term. This demonstrates how to find and simplify the middle term of a binomial expansion.
Given expansion is \( (1 + x)^{2n} \).
Here, the power is \( N = 2n \). Since \( N \) is even, there is one middle term.
The middle term is \( t_{\frac{2n}{2}+1} = t_{n+1} \).
For \( t_{n+1} \), we have \( r=n \).
Using the general term formula \( t_{r+1} = N C_r X^{N-r} A^r \):
\( t_{n+1} = (2nC_n) (1)^{2n-n} (x)^n \)
\( t_{n+1} = (2nC_n) \cdot 1^n \cdot x^n \)
\( t_{n+1} = (2nC_n) x^n \)
From the proof in Question 6, we know that \( (2nC_n) = \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) 2^n}{n!} \).
Substituting this expression back:
\( t_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1) 2^n}{n!} x^n \)
Hence proved.
In simple words: To find the middle term of \( (1+x)^{2n} \), we use the middle term formula, which tells us it's \( t_{n+1} \). Then we plug in the values for \( x \) and \( a \) and simplify the combination part using what we learned in the last question. This leads us to the final answer that includes \( x^n \).

🎯 Exam Tip: When proving properties of middle terms, clearly state if `n` is even or odd, as this determines the number of middle terms. Leverage previously proven results, like the expansion of `(2n)C_n`, to simplify the current proof.