RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.4

Question 1. Expand the following Binomials upto fourth term:
(i) \( (1 + x^2)^{-2} \)
(ii) \( \left(1 - \frac{x}{2}\right)^{1/2} \)
(iii) \( (3 - 2x^2)^{-2/3} \)
(iv) \( \frac{1}{\sqrt{5+4x}} \)
Answer:
(i) Expansion upto four terms of \( (1 + x^2)^{-2} \):
We use the formula for binomial expansion: \( (1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + ... \)
Here, \( n = -2 \) and \( z = x^2 \).
\( (1+x^2)^{-2} = 1 + (-2)x^2 + \frac{(-2)(-2-1)}{2!}(x^2)^2 + \frac{(-2)(-2-1)(-2-2)}{3!}(x^2)^3 + ... \)
\( = 1 - 2x^2 + \frac{(-2)(-3)}{2}(x^4) + \frac{(-2)(-3)(-4)}{6}(x^6) + ... \)
\( = 1 - 2x^2 + 3x^4 - 4x^6 \)

(ii) Expansion upto four terms of \( \left(1 - \frac{x}{2}\right)^{1/2} \):
Here, \( n = \frac{1}{2} \) and \( z = -\frac{x}{2} \).
\( \left(1 - \frac{x}{2}\right)^{1/2} = 1 + \left(\frac{1}{2}\right)\left(-\frac{x}{2}\right) + \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}\left(-\frac{x}{2}\right)^2 + \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}\left(-\frac{x}{2}\right)^3 + ... \)
\( = 1 - \frac{x}{4} + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}\left(\frac{x^2}{4}\right) + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6}\left(-\frac{x^3}{8}\right) + ... \)
\( = 1 - \frac{x}{4} - \frac{1}{8}\left(\frac{x^2}{4}\right) + \frac{3}{48}\left(-\frac{x^3}{8}\right) + ... \)
\( = 1 - \frac{x}{4} - \frac{x^2}{32} - \frac{x^3}{128} \)

(iii) Expansion upto four terms of \( (3 - 2x^2)^{-2/3} \):
First, factor out 3: \( (3 - 2x^2)^{-2/3} = 3^{-2/3}\left(1 - \frac{2x^2}{3}\right)^{-2/3} \)
Now, use the binomial expansion with \( n = -\frac{2}{3} \) and \( z = -\frac{2x^2}{3} \).
\( = 3^{-2/3} \left[ 1 + \left(-\frac{2}{3}\right)\left(-\frac{2x^2}{3}\right) + \frac{\left(-\frac{2}{3}\right)\left(-\frac{2}{3}-1\right)}{2!}\left(-\frac{2x^2}{3}\right)^2 + \frac{\left(-\frac{2}{3}\right)\left(-\frac{2}{3}-1\right)\left(-\frac{2}{3}-2\right)}{3!}\left(-\frac{2x^2}{3}\right)^3 + ... \right] \)
\( = 3^{-2/3} \left[ 1 + \frac{4x^2}{9} + \frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{2}\left(\frac{4x^4}{9}\right) + \frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{6}\left(-\frac{8x^6}{27}\right) + ... \right] \)
\( = 3^{-2/3} \left[ 1 + \frac{4x^2}{9} + \frac{10}{9 \times 2}\left(\frac{4x^4}{9}\right) + \frac{-80}{27 \times 6}\left(-\frac{8x^6}{27}\right) + ... \right] \)
\( = 3^{-2/3} \left[ 1 + \frac{4x^2}{9} + \frac{20x^4}{81} + \frac{320x^6}{2187} + ... \right] \)

(iv) Expansion upto four terms of \( \frac{1}{\sqrt{5+4x}} \):
Rewrite the expression: \( \frac{1}{\sqrt{5+4x}} = (5+4x)^{-1/2} \)
Factor out 5: \( (5+4x)^{-1/2} = 5^{-1/2}\left(1 + \frac{4x}{5}\right)^{-1/2} \)
Now, use the binomial expansion with \( n = -\frac{1}{2} \) and \( z = \frac{4x}{5} \).
\( = 5^{-1/2} \left[ 1 + \left(-\frac{1}{2}\right)\left(\frac{4x}{5}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2!}\left(\frac{4x}{5}\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}\left(\frac{4x}{5}\right)^3 + ... \right] \)
\( = 5^{-1/2} \left[ 1 - \frac{2x}{5} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\frac{16x^2}{25}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}\left(\frac{64x^3}{125}\right) + ... \right] \)
\( = 5^{-1/2} \left[ 1 - \frac{2x}{5} + \frac{3}{8}\left(\frac{16x^2}{25}\right) - \frac{15}{48}\left(\frac{64x^3}{125}\right) + ... \right] \)
\( = \frac{1}{\sqrt{5}} \left[ 1 - \frac{2x}{5} + \frac{6x^2}{25} - \frac{4x^3}{25} + ... \right] \)
In simple words: To expand a binomial up to a certain number of terms, use the general binomial theorem formula. Replace 'n' with the power and 'z' with the second term in the binomial. Calculate each term carefully, paying attention to signs and fractions.

🎯 Exam Tip: Remember to factor out the first term if it's not 1 before applying the binomial expansion formula, especially for expressions like \( (a+b)^n \).

Question 2. Find the required terms in the following expansions:
(i) Fourth term of \( (1 – 3x)^{-1/3} \)
(ii) Seventh term of \( (1 + x)^{5/2} \)
(iii) Eighth term of \( (1 + 2x)^{-1/2} \)
Answer:
We use the general term formula for binomial expansion: \( T_{r+1} = \frac{n(n-1)...(n-r+1)}{r!} z^r \).

(i) Fourth term of \( (1 – 3x)^{-1/3} \):
Here, \( n = -\frac{1}{3} \), \( z = -3x \), and for the fourth term, \( r+1 = 4 \implies r = 3 \).
\( T_4 = \frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)\left(-\frac{1}{3}-2\right)}{3!}(-3x)^3 \)
\( = \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right)}{6}(-27x^3) \)
\( = \frac{-\frac{28}{27}}{6}(-27x^3) \)
\( = \left(-\frac{28}{27 \times 6}\right)(-27x^3) \)
\( = \frac{28}{6}x^3 = \frac{14}{3}x^3 \)

(ii) Seventh term of \( (1 + x)^{5/2} \):
Here, \( n = \frac{5}{2} \), \( z = x \), and for the seventh term, \( r+1 = 7 \implies r = 6 \).
\( T_7 = \frac{\frac{5}{2}\left(\frac{5}{2}-1\right)\left(\frac{5}{2}-2\right)\left(\frac{5}{2}-3\right)\left(\frac{5}{2}-4\right)\left(\frac{5}{2}-5\right)}{6!}x^6 \)
\( = \frac{\frac{5}{2}\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{720}x^6 \)
\( = \frac{\frac{5 \times 3 \times 1 \times (-1) \times (-3) \times (-5)}{64}}{720}x^6 \)
\( = \frac{-\frac{225}{64}}{720}x^6 \)
\( = -\frac{225}{64 \times 720}x^6 \)
\( = -\frac{5 \times 45}{64 \times 16 \times 45}x^6 \)
\( = -\frac{5}{1024}x^6 \)

(iii) Eighth term of \( (1 + 2x)^{-1/2} \):
Here, \( n = -\frac{1}{2} \), \( z = 2x \), and for the eighth term, \( r+1 = 8 \implies r = 7 \).
\( T_8 = \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\left(-\frac{1}{2}-3\right)\left(-\frac{1}{2}-4\right)\left(-\frac{1}{2}-5\right)\left(-\frac{1}{2}-6\right)}{7!}(2x)^7 \)
\( = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{7}{2}\right)\left(-\frac{9}{2}\right)\left(-\frac{11}{2}\right)\left(-\frac{13}{2}\right)}{5040}(128x^7) \)
\( = \frac{(-1)^7 (1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13)}{2^7 \cdot 7!} (2^7 x^7) \)
\( = \frac{-13 \times 11 \times 9 \times 7 \times 5 \times 3 \times 1}{5040 \times 128} (128x^7) \)
\( = -\frac{13 \times 11 \times 9 \times 7 \times 5 \times 3}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}x^7 \)
\( = -\frac{13 \times 11 \times 9}{6 \times 4 \times 2}x^7 \)
\( = -\frac{1287}{48}x^7 = -\frac{429}{16}x^7 \)
In simple words: To find a specific term in an expansion, use the general term formula. Make sure you correctly identify 'n' (the power), 'z' (the second part of the binomial), and 'r' (which is one less than the term number you want).

🎯 Exam Tip: Be very careful with the signs and fractions in the numerator of the general term formula. A single mistake can lead to an incorrect result.

Question 3. Find the general term of the following expansions:
(i) \( (a^3 - x^3)^{2/3} \)
(ii) \( (1 - 2x)^{-3/2} \)
(iii) \( (1 – x)^{-p/q} \)
Answer:
The general term \( T_{r+1} \) for \( (1+z)^n \) is \( \frac{n(n-1)...(n-r+1)}{r!} z^r \).

(i) For \( (a^3 - x^3)^{2/3} \):
First, factor out \( a^3 \): \( (a^3 - x^3)^{2/3} = (a^3)^{2/3}\left(1 - \frac{x^3}{a^3}\right)^{2/3} = a^2\left(1 - \frac{x^3}{a^3}\right)^{2/3} \)
Here, \( n = \frac{2}{3} \) and \( z = -\frac{x^3}{a^3} \).
\( T_{r+1} = a^2 \frac{\frac{2}{3}\left(\frac{2}{3}-1\right)...\left(\frac{2}{3}-r+1\right)}{r!}\left(-\frac{x^3}{a^3}\right)^r \)
\( = a^2 (-1)^r \frac{\frac{2}{3}\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)...\left(\frac{2}{3}-r+1\right)}{r!} \frac{x^{3r}}{a^{3r}} \)
\( = a^2 (-1)^r \frac{2 \cdot 1 \cdot 4 \cdot ... \cdot (3r-5)}{3^r \cdot r!} \frac{x^{3r}}{a^{3r}} \)
\( = (-1)^r \frac{2 \cdot 1 \cdot 4 \cdot ... \cdot (3r-5)}{3^r \cdot r!} \frac{x^{3r}}{a^{3r-2}} \)

(ii) For \( (1 - 2x)^{-3/2} \):
Here, \( n = -\frac{3}{2} \) and \( z = -2x \).
\( T_{r+1} = \frac{\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-1\right)...\left(-\frac{3}{2}-r+1\right)}{r!}(-2x)^r \)
\( = \frac{\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)...\left(-\frac{2r+1}{2}\right)}{r!}(-1)^r 2^r x^r \)
\( = \frac{(-1)^r (3 \cdot 5 \cdot 7 \cdot ... \cdot (2r+1))}{2^r \cdot r!}(-1)^r 2^r x^r \)
\( = \frac{(-1)^{2r} (3 \cdot 5 \cdot 7 \cdot ... \cdot (2r+1))}{r!} x^r \)
Since \( (-1)^{2r} = 1 \),
\( T_{r+1} = \frac{3 \cdot 5 \cdot 7 \cdot ... \cdot (2r+1)}{r!} x^r \)

(iii) For \( (1 – x)^{-p/q} \):
Here, \( n = -\frac{p}{q} \) and \( z = -x \).
\( T_{r+1} = \frac{\left(-\frac{p}{q}\right)\left(-\frac{p}{q}-1\right)...\left(-\frac{p}{q}-r+1\right)}{r!}(-x)^r \)
\( = \frac{(-1)^r \frac{p}{q}\left(\frac{p}{q}+1\right)...\left(\frac{p}{q}+r-1\right)}{r!}(-1)^r x^r \)
\( = \frac{(-1)^{2r} p(p+q)(p+2q)...(p+(r-1)q)}{q^r \cdot r!} x^r \)
\( = \frac{p(p+q)(p+2q)...(p+(r-1)q)}{q^r \cdot r!} x^r \)
In simple words: The general term (the \( r+1 \)th term) shows the pattern for any term in the expansion. To find it, substitute the specific 'n' (the power) and 'z' (the variable part) into the general formula, and simplify the expression.

🎯 Exam Tip: When finding the general term, simplify the product of 'n' terms in the numerator carefully. Group similar factors and powers of -1 or any constant.

Question 4. If x < 3, find the coefficient of \( x^5 \) in the expansion of \( (3 - x)^{-8} \).
Answer:
The expansion is \( (3 - x)^{-8} \). First, factor out 3:
\( (3 - x)^{-8} = 3^{-8}\left(1 - \frac{x}{3}\right)^{-8} \)
Now, we use the general term formula for \( (1+z)^n \), where \( n = -8 \) and \( z = -\frac{x}{3} \).
The \( r+1 \)th term is \( T_{r+1} = 3^{-8} \frac{n(n-1)...(n-r+1)}{r!} z^r \)
\( T_{r+1} = 3^{-8} \frac{(-8)(-8-1)...(-8-r+1)}{r!} \left(-\frac{x}{3}\right)^r \)
\( = 3^{-8} \frac{(-8)(-9)...(-(7+r))}{r!} (-1)^r \frac{x^r}{3^r} \)
\( = 3^{-8} (-1)^r \frac{(8)(9)...(7+r)}{r!} (-1)^r \frac{x^r}{3^r} \)
\( = \frac{(8)(9)...(7+r)}{r! 3^{8+r}} x^r \)
We need the coefficient of \( x^5 \), so we set \( r = 5 \).
The coefficient of \( x^5 \) is \( \frac{(8)(9)(10)(11)(12)}{5! 3^{8+5}} \)
\( = \frac{8 \times 9 \times 10 \times 11 \times 12}{5 \times 4 \times 3 \times 2 \times 1 \times 3^{13}} \)
\( = \frac{95040}{120 \times 3^{13}} = \frac{792}{3^{13}} \)
\( = \frac{792}{1594323} \)
To simplify, \( 792 = 8 \times 9 \times 11 = 2^3 \times 3^2 \times 11 \).
So, \( \frac{2^3 \times 3^2 \times 11}{3^{13}} = \frac{2^3 \times 11}{3^{11}} = \frac{8 \times 11}{177147} = \frac{88}{177147} \)
In simple words: To find the part with \( x^5 \), first make the starting part of the binomial 1. Then use the general term formula. Set the power of x to 5 to find which term it is. Finally, calculate the number that comes with \( x^5 \).

🎯 Exam Tip: Always make sure the first term of the binomial is 1 before applying the standard binomial expansion formulas. Factor out the constant if needed.

Question 5. Find the coefficient of \( x^6 \) in the expansion of \( (a + 2bx^2)^{-3} \).
Answer:
The expansion is \( (a + 2bx^2)^{-3} \). Factor out 'a' first:
\( (a + 2bx^2)^{-3} = a^{-3}\left(1 + \frac{2bx^2}{a}\right)^{-3} \)
Now, we use the general term formula, where \( n = -3 \) and \( z = \frac{2bx^2}{a} \).
The \( r+1 \)th term is \( T_{r+1} = a^{-3} \frac{n(n-1)...(n-r+1)}{r!} z^r \)
\( T_{r+1} = a^{-3} \frac{(-3)(-3-1)...(-3-r+1)}{r!} \left(\frac{2bx^2}{a}\right)^r \)
\( = a^{-3} \frac{(-3)(-4)...(-(r+2))}{r!} \frac{2^r b^r x^{2r}}{a^r} \)
\( = a^{-3} (-1)^r \frac{3 \cdot 4 \cdot ... \cdot (r+2)}{r!} \frac{2^r b^r x^{2r}}{a^r} \)
\( = (-1)^r \frac{3 \cdot 4 \cdot ... \cdot (r+2) \cdot 2^r b^r}{r! a^{r+3}} x^{2r} \)
We need the coefficient of \( x^6 \), so we set the power of x to 6:
\( 2r = 6 \implies r = 3 \).
Now substitute \( r=3 \) into the coefficient part of the general term:
Coefficient of \( x^6 = (-1)^3 \frac{3 \cdot 4 \cdot 5}{3!} \frac{2^3 b^3}{a^{3+3}} \)
\( = -1 \cdot \frac{60}{6} \cdot \frac{8b^3}{a^6} \)
\( = -1 \cdot 10 \cdot \frac{8b^3}{a^6} \)
\( = -\frac{80b^3}{a^6} \)
In simple words: To find the number that multiplies \( x^6 \), first rewrite the expression so that the first part inside the bracket is 1. Then, use the general term formula. Since the x term is \( x^2 \), you need to find 'r' such that \( 2r = 6 \). Once you have 'r', put it back into the formula to get the coefficient.

🎯 Exam Tip: When the variable term itself has a power (e.g., \( x^2 \)), make sure to set the exponent of the variable from the general term formula (e.g., \( 2r \)) equal to the desired power (e.g., 6).

Question 6. Find the coefficient of \( x^{10} \) in the expansion of \( \frac{1+3x^2}{(1-x^2)^3} \).
Answer:
The given expression is \( (1+3x^2)(1-x^2)^{-3} \).
We need to expand \( (1-x^2)^{-3} \) using the binomial theorem \( (1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + ... \).
Here, \( n = -3 \) and \( z = -x^2 \).
\( (1-x^2)^{-3} = 1 + (-3)(-x^2) + \frac{(-3)(-3-1)}{2!}(-x^2)^2 + \frac{(-3)(-3-1)(-3-2)}{3!}(-x^2)^3 + \frac{(-3)(-3-1)(-3-2)(-3-3)}{4!}(-x^2)^4 + \frac{(-3)(-3-1)(-3-2)(-3-3)(-3-4)}{5!}(-x^2)^5 + ... \)
\( = 1 + 3x^2 + \frac{(-3)(-4)}{2}x^4 + \frac{(-3)(-4)(-5)}{6}(-x^6) + \frac{(-3)(-4)(-5)(-6)}{24}x^8 + \frac{(-3)(-4)(-5)(-6)(-7)}{120}(-x^{10}) + ... \)
\( = 1 + 3x^2 + 6x^4 + 10x^6 + 15x^8 + 21x^{10} + ... \)
Now, we multiply this expansion by \( (1+3x^2) \):
\( (1+3x^2)(1-x^2)^{-3} = (1+3x^2)(1 + 3x^2 + 6x^4 + 10x^6 + 15x^8 + 21x^{10} + ...) \)
To find the coefficient of \( x^{10} \), we look for terms that multiply to give \( x^{10} \):
1. From \( 1 \) (first term of \( 1+3x^2 \)): It multiplies with \( 21x^{10} \) from the expansion, giving \( 1 \times 21x^{10} = 21x^{10} \).
2. From \( 3x^2 \) (second term of \( 1+3x^2 \)): It multiplies with \( 15x^8 \) from the expansion, giving \( 3x^2 \times 15x^8 = 45x^{10} \).
The total coefficient of \( x^{10} \) is the sum of these:
\( 21 + 45 = 66 \).
In simple words: First, expand the fraction part \( (1-x^2)^{-3} \) into a series of terms. Then, multiply this series by \( (1+3x^2) \). Look for all pairs of terms (one from \( (1+3x^2) \) and one from the series) that, when multiplied, give \( x^{10} \). Add up the numbers from these pairs to get the final answer.

🎯 Exam Tip: When multiplying two series to find a specific coefficient, list out the terms from each series that will combine to form the desired power. This systematic approach reduces errors.

Question 7. Find the coefficient of x in the expansion of \( (1 – 2x + 3x^2 - 4x^3 + ...) \) and if \( x = \frac{1}{2} \) and \( n = 1 \), then find the value of the expression.
Answer:
The given series is \( (1 – 2x + 3x^2 - 4x^3 + ...) \).
This is a known binomial expansion for \( (1+x)^{-2} \).
So, \( (1 – 2x + 3x^2 - 4x^3 + ...) = (1+x)^{-2} \).
We need the coefficient of x in this expansion. The coefficient of x in \( (1+x)^{-2} \) is \( -2 \).

Now, the full expression is \( (1 – 2x + 3x^2 - 4x^3 + ...)^n \).
Substituting the recognized form, this becomes \( ((1+x)^{-2})^n = (1+x)^{-2n} \).
We are given \( x = \frac{1}{2} \) and \( n = 1 \).
Substitute these values into the simplified expression \( (1+x)^{-2n} \):
\( \left(1 + \frac{1}{2}\right)^{-2(1)} \)
\( = \left(\frac{3}{2}\right)^{-2} \)
\( = \left(\frac{2}{3}\right)^2 \)
\( = \frac{4}{9} \)
In simple words: First, notice that the series \( (1 - 2x + 3x^2 - ...) \) is the same as \( (1+x)^{-2} \). The coefficient of x in this series is -2. Then, substitute this back into the original expression, which becomes \( (1+x)^{-2n} \). Finally, put in the given values for x and n and calculate the result.

🎯 Exam Tip: It is crucial to recognize common series expansions like \( (1+x)^{-1} \), \( (1-x)^{-1} \), \( (1+x)^{-2} \), and \( (1-x)^{-2} \) as they frequently appear in problems.

Question 9. Prove that \( (1 + x + x^2 + x^3 +....) (1 + 3x + 6x^2+ ...) = (1 + 2x + 3x^2 + ...)2 \).
Answer:
We know the following standard binomial expansions:
1. \( (1-x)^{-1} = 1 + x + x^2 + x^3 + ... \)
2. \( (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + ... \)
3. \( (1-x)^{-3} = 1 + 3x + 6x^2 + 10x^3 + ... \)

Now, let's look at the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = (1 + x + x^2 + x^3 +....) (1 + 3x + 6x^2+ ...) \)
Substitute the known forms:
L.H.S. \( = (1-x)^{-1} \cdot (1-x)^{-3} \)
Using the rule \( a^m \cdot a^n = a^{m+n} \):
L.H.S. \( = (1-x)^{-1-3} \)
L.H.S. \( = (1-x)^{-4} \)

Now, let's look at the Right Hand Side (R.H.S.) of the equation:
R.H.S. \( = (1 + 2x + 3x^2 + ...)^2 \)
Substitute the known form:
R.H.S. \( = ((1-x)^{-2})^2 \)
Using the rule \( (a^m)^n = a^{mn} \):
R.H.S. \( = (1-x)^{-2 \times 2} \)
R.H.S. \( = (1-x)^{-4} \)
Since L.H.S. \( = (1-x)^{-4} \) and R.H.S. \( = (1-x)^{-4} \),
L.H.S. = R.H.S.
Hence Proved.
In simple words: We can prove this by recognizing that each long series is actually a simpler binomial expression. The first series is \( (1-x)^{-1} \), the second is \( (1-x)^{-3} \), and the third is \( (1-x)^{-2} \). By replacing the series with these simpler forms, both sides of the equation simplify to the same expression, proving they are equal.

🎯 Exam Tip: Memorizing the expansions of \( (1-x)^{-1}, (1-x)^{-2}, \) and \( (1-x)^{-3} \) is very useful for quickly solving problems involving infinite series proofs.

Question 10. If \( x = 2y + 3y^2 + 4y^3 + ... \) then express y in series of as ascending powers of x.
Answer:
Given the series: \( x = 2y + 3y^2 + 4y^3 + ... \)
We can rewrite this series by adding and subtracting 'y':
\( x = (y + 2y + 3y^2 + 4y^3 + ...) - y \)
This can be recognized as a known series:
We know that \( (1-y)^{-2} = 1 + 2y + 3y^2 + 4y^3 + ... \)
So, the expression for x becomes:
\( x = ( (1-y)^{-2} - 1 ) \)
Now, we want to express y in terms of x. Let's rearrange the equation:
\( x = (1-y)^{-2} - 1 \)
Add 1 to both sides:
\( 1 + x = (1-y)^{-2} \)
Take the square root of both sides:
\( \sqrt{1+x} = (1-y)^{-1} \)
Since \( (1-y)^{-1} = \frac{1}{1-y} \), we have:
\( \sqrt{1+x} = \frac{1}{1-y} \)
Now, swap the terms to isolate \( (1-y) \):
\( 1-y = \frac{1}{\sqrt{1+x}} \)
\( 1-y = (1+x)^{-1/2} \)
To find y, subtract 1 from both sides and multiply by -1 (or move y to one side and the rest to the other):
\( y = 1 - (1+x)^{-1/2} \)
Now, expand \( (1+x)^{-1/2} \) using the binomial theorem \( (1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + ... \)
Here, \( n = -\frac{1}{2} \) and \( z = x \).
\( (1+x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2!}x^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}x^3 + ... \)
\( = 1 - \frac{1}{2}x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}x^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}x^3 + ... \)
\( = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{15}{48}x^3 + ... \)
\( = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + ... \)
Substitute this back into the equation for y:
\( y = 1 - \left(1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + ...\right) \)
\( y = 1 - 1 + \frac{1}{2}x - \frac{3}{8}x^2 + \frac{5}{16}x^3 - ... \)
\( y = \frac{1}{2}x - \frac{3}{8}x^2 + \frac{5}{16}x^3 - ... \)
In simple words: The given equation for x looks like part of a known binomial series for \( (1-y)^{-2} \). By carefully rearranging the formula and taking the square root, we can find \( (1-y)^{-1} \). Then, by expressing y in terms of x and expanding \( (1+x)^{-1/2} \), we can write y as a series of powers of x.

🎯 Exam Tip: Problems requiring expressing one variable in terms of another often involve recognizing a known series, algebraically manipulating the equation, and then applying binomial expansion again.