CBSE Class 11 Mathematics Binomial Theorem Worksheet Set 04

CBSE Class 11 Mathematics Worksheet - Binomial Theorem (3). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. The \(2^{\text{nd}}, 3^{\text{rd}} \ \& \ 4^{\text{th}}\) terms in the expansion of \( (x + a)^n \) are 240, 720 & 1080. Find \(x\), \(a\) & \(n\)?
Answer: Given expansion: \( (x + a)^n \)
\( T_2 = 240, T_3 = 720 \ \& \ T_4 = 1080 \)
General term: \( T_{r+1} = {}^nC_r x^{n-r} a^r \)
Now, \( T_2 = {}^nC_1 x^{n-1} a^1 = 240 \) ............. (1)
\( = T_3 = {}^nC_2 x^{n-2} a^2 = 720 \) ............. (2)
\( = T_4 = {}^nC_3 x^{n-3} a^3 = 1080 \) ............. (3)
Equation (2) \( \div \) equation (1)
\( \Rightarrow \frac{{}^nC_2 x^{n-2} a^2}{{}^nC_1 x^{n-1} a^1} = \frac{720}{240} = 3 \)
\( \Rightarrow \frac{\frac{n(n-1)}{2} \cdot x^{n-2-n+1} \cdot a}{n} = 3 \)
\( \Rightarrow \frac{(n-1)x^{-1}a}{2} = 3 \)
\( \Rightarrow \frac{(n-1)a}{2x} = 3 \)
\( \Rightarrow (n - 1)a = 6x \) ............. (4)
Now, equation (3) \( \div \) equation (2)
\( \Rightarrow \frac{{}^nC_3 x^{n-3} a^3}{{}^nC_2 x^{n-2} a^2} = \frac{1080}{720} \)
\( \Rightarrow \frac{\frac{n(n-1)(n-2)}{6} \cdot x^{n-3-n+2} \cdot a}{\frac{n(n-1)}{2}} = \frac{3}{2} \)
\( \Rightarrow \frac{(n-2) \cdot x^{-1} a}{3} = \frac{3}{2} \)
\( \Rightarrow \frac{(n-2)a}{3x} = \frac{3}{2} \)
\( \Rightarrow 2(n - 2)a = 9x \) .................... (5)
Now, equation (5) \( \div \) equation (4)
\( \Rightarrow \frac{2(n-2)a}{(n-1)a} = \frac{9x}{6x} \)
\( \Rightarrow \frac{2n-4}{n-1} = \frac{3}{2} \)
\( \Rightarrow 4n-8 = 3n-3 \)
\( \Rightarrow n=5 \), put this value in equation (4)
We get, \( 4a=6x \)
\( \Rightarrow a = \frac{3x}{2} \), put value of \(n\) & \(a\) in equation (1)
We have, \( {}^5C_1 (x)^4 \cdot \left(\frac{3x}{2}\right) = 240 \)
\( \Rightarrow 5x^4 \cdot \frac{3x}{2} = 240 \)
\( \Rightarrow x^5 = \frac{240 \times 2}{15} \)
\( \Rightarrow x^5 = 32 = 2^5 \)
\( \Rightarrow x = 2, a = 3 \dots\dots\dots\dots \left\{ \text{since } a = \frac{3x}{2} \right\} \)
\( \therefore n = 5, x = 2 \text{ and } a = 3 \) ans.

Question. Find \(a\), \(b\) & \(n\) in expansion of \( (a + b)^n \), if the first three terms in the expansion are 729, 7290 & 30375?
Answer: Given expansion: \( (a + b)^n \)
\( T_1 = 729, T_2 = 7290 \ \& \ T_3 = 30375 \)
General terms : \( T_{r+1} = {}^nC_r a^{n-r} b^r \)
Now, \( T_1 = {}^nC_0 a^n b^0 = 729 \)
\( \Rightarrow T_1 = a^n = 729 \) ............. (1)
\( \Rightarrow T_2 = {}^nC_1 a^{n-1} b^1 = 7290 \) ............. (2)
\( \Rightarrow T_3 = {}^nC_2 a^{n-2} b^2 = 30375 \) ............. (3)
Now, equation (2) \( \div \) equation (1)
\( \Rightarrow \frac{{}^nC_1 a^{n-1} b^1}{a^n} = \frac{7290}{729} \)
\( \Rightarrow n \cdot a^{n-1-n} \cdot b = 10 \)
\( \Rightarrow \frac{nb}{a} = 10 \)
\( \Rightarrow nb = 10a \) ...................... (4)
Now, equation (3) \( \div \) equation (2)
\( \Rightarrow \frac{{}^nC_2 a^{n-2} b^2}{{}^nC_1 a^{n-1} b^1} = \frac{30375}{7290} \)
\( \Rightarrow \frac{\frac{n(n-1)}{2} \cdot a^{n-2-n+1} \cdot b}{n} = \frac{25}{6} \)
\( \Rightarrow \frac{(n-1) \cdot a^{-1} \cdot b}{2} = \frac{25}{6} \)
\( \Rightarrow \frac{(n-1)b}{2a} = \frac{25}{6} \)
\( \Rightarrow 6(n - 1)b = 50a \) ............. (5)
Now, equation (5) \( \div \) (4)
\( \Rightarrow \frac{6(n-1)b}{nb} = \frac{50a}{10a} \)
\( \Rightarrow 6n - 6 = 5n \)
\( \Rightarrow n = 6 \), put the value of \(n\) in equation (4)
\( \Rightarrow 6b = 10a \)
\( \Rightarrow b = \frac{5a}{3} \)
Now, from equation (1) put \(n=6\)
\( \Rightarrow a^6 = 729 \)
\( \Rightarrow a^6 = 3^6 \)
\( \Rightarrow a = 3 \)
\( \therefore n = 6, a = 3 \ \& \ b = 5 \) ans.

Question. If the coefficient of \( a^{r-1} \), \( a^r \) and \( a^{r+1} \) in the expansion of \( (1 + a)^n \) are in A.P, show that \( n^2 - n(4r + 1) + 4r^2 - 2 = 0 \)?
Answer: Given expansion: \( (1 + a)^n \)
General term: \( T_{r+1} = {}^nC_r (1)^{n-r} a^r \)
\( T_{r+1} = {}^nC_r a^r \)
Clearly, coefficient of \( a^r = {}^nC_r \)
\( \therefore \) coefficient of \( a^{r-1} = {}^nC_{r-1} \)
And coefficient of \( a^{r+1} = {}^nC_{r+1} \)
We are given that, \( {}^nC_{r-1} \), \( {}^nC_r \) and \( {}^nC_{r+1} \) are in A.P.
\( \Rightarrow 2 \cdot {}^nC_r = {}^nC_{r-1} + {}^nC_{r+1} \)
\( \Rightarrow 2 \cdot \frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!} + \frac{n!}{(r+1)!(n-r-1)!} \)
\( \Rightarrow \frac{2}{r!(n-r)!} = \frac{1}{(r-1)!(n-r+1)!} + \frac{1}{(r+1)!(n-r-1)!} \)
\( \Rightarrow \frac{2}{r(r-1)!(n-r-1)!} = \frac{1}{(r-1)!(n-r+1)(n-r)(n-r-1)!} + \frac{1}{(r+1)r!(r-1)(n-r+1)!} \)
\( \Rightarrow \frac{2}{r(n-r)} - \frac{1}{(n-r+1)(n-r)} = \frac{1}{(r+1)r} \)
\( \Rightarrow \frac{2(n-r+1)-r}{r(n-r)(n-r+1)} = \frac{1}{r+1} \)
\( \Rightarrow 2nr + 2n - 3r^2 - 3r + 2r + 2 = n^2 - nr + n - nr + r^2 - r \)
\( \Rightarrow n^2 - 4nr - n + 4r^2 - 2 = 0 \)
\( \Rightarrow n^2 - n(4r + 1) + 4r^2 - 2 = 0 \) (proved)

Question. Find the 4th term from the end in the expansion of \( \left(x^4 - \frac{1}{x^3}\right)^{11} \).
Answer: Given expansion : \( \left(x^4 - \frac{1}{x^3}\right)^{11} \)
General terms: \( T_{r+1} = (-1)^r {}^{11}C_r (x^4)^{11-r} \left(\frac{1}{x^3}\right)^r \)
\( \Rightarrow T_{r+1} = (-1)^r {}^{11}C_r (x)^{44-4r} \cdot \frac{1}{x^{3r}} \)
\( \Rightarrow T_{r+1} = (-1)^r {}^{11}C_r (x)^{44-7r} \)
Formula, \( r^{\text{th}} \) term from the end = \( (n - 1 + 2)^{\text{th}} \) term from beginning and \( 4^{\text{th}} \) term from the end = \( (11 - 4 + 2)^{\text{th}} \) term from beginning = \( 9^{\text{th}} \) term.
For \( T_9 \), put \( r = 8 \)
\( \Rightarrow T_9 = (-1)^8 {}^{11}C_8 (x)^{44-56} \)
\( \Rightarrow T_9 = {}^{11}C_3 (x)^{-12} \dots\dots\dots\dots \{ {}^nC_r = {}^nC_{n-r} \} \)
\( \Rightarrow T_9 = \frac{11 \times 10 \times 9}{6} \cdot \frac{1}{x^{12}} \)
\( \Rightarrow T_9 = \frac{165}{x^{12}} \)
\( \therefore 4^{\text{th}} \) term from the end = \( \frac{165}{x^{12}} \) ans.

Question. Find the value of \(n\), if the ratio of the 5th term from the beginning to the 5th term from the end in the expansion of \( \left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}}\right)^n \) is \( \sqrt{6} : 1 \).
Answer: Expansion: \( \left(2^{1/4} + \frac{1}{3^{1/4}}\right)^n \)
General term: \( T_{r+1} = {}^nC_r \left(2^{1/4}\right)^{n-r} \left(\frac{1}{3^{1/4}}\right)^r \)
\( \Rightarrow T_{r+1} = {}^nC_r (2)^{\frac{n-r}{4}} \cdot \frac{1}{3^{\frac{r}{4}}} \)
5th term from the beginning, put \(r = 4\)
\( \Rightarrow T_5 = {}^nC_4 (2)^{\frac{n-4}{4}} \cdot \frac{1}{3} \)
Now, 5th term from the end = \( (n - 5 + 2)^{\text{th}} \) term from the beginning = \( (n - 3)^{\text{rd}} \) term
For \(T_{n-3}\), put \(r = n-4\)
\( \Rightarrow T_{n-3} = {}^nC_{n-4} (2)^{\frac{n-(n-4)}{4}} \cdot \frac{1}{3^{\frac{n-4}{4}}} \)
\( \Rightarrow T_{n-3} = {}^nC_{n-4} (2)^1 \cdot \frac{1}{3^{\frac{n-4}{4}}} \)
Given, \( \frac{T_5}{T_{n-3}} = \frac{\sqrt{6}}{1} \)
\( \Rightarrow \frac{{}^nC_4 (2)^{\frac{n-4}{4}} \cdot \frac{1}{3}}{{}^nC_{n-4} (2)^1 \cdot \frac{1}{3^{\frac{n-4}{4}}}} = \frac{\sqrt{6}}{1} \)
\( \Rightarrow \frac{\frac{n!}{4!(n-4)!} \cdot (2)^{\frac{n-4}{4}-1} \cdot 3^{\frac{n-4}{4}}}{\frac{n!}{(n-4)!4!} \cdot 3} = \frac{\sqrt{6}}{1} \)
\( \Rightarrow (2)^{\frac{n-8}{4}} \cdot (3)^{\frac{n-8}{4}} = \sqrt{6} \)
\( \Rightarrow (6)^{\frac{n-8}{4}} = (6)^{\frac{1}{2}} \)
\( \Rightarrow \frac{n-8}{4} = \frac{1}{2} \)
\( \Rightarrow 2n - 16 = 4 \)
\( \Rightarrow 2n = 20 \)
\( \Rightarrow n = 10 \) ans.

Question. Prove that there is no term including \(x^6\) in the expansion of \( \left(2x^2 - \frac{3}{x}\right)^{11} \)?
Answer: General terms: \( T_{r+1} = (-1)^r {}^{11}C_r \left(2x^2\right)^{11-r} \left(\frac{3}{x}\right)^r \)
\( \Rightarrow T_{r+1} = (-1)^r {}^{11}C_r (2)^{11-r} \cdot (x)^{22-2r} \cdot \frac{3^r}{x^r} \)
\( \Rightarrow T_{r+1} = (-1)^r {}^{11}C_r (2)^{11-r} \cdot (3)^r \cdot (x)^{22-3r} \)
Let \(x^6\) occurs in the \((r + 1)^{\text{th}}\) term then, for \(x^6\) put \(22 - 3r = 6\)
\( \Rightarrow 3r = 16 \)
\( \Rightarrow r = \frac{16}{3} \), which is in fraction but ‘r’ cannot be in fraction or negative
\( \therefore \) there is no term in the expansion which involves \(x^6\) ans.

Question. If the 4th term in the expansion of \( \left(ax - \frac{1}{x}\right)^n \) is \( \frac{5}{2} \), then find the values of ‘a’ & ‘n’?
Answer: Expansion: \( \left(ax - \frac{1}{x}\right)^n \)
General term: \( T_{r+1} = {}^nC_r (ax)^{n-r} \left(\frac{1}{x}\right)^r \)
\( \Rightarrow T_{r+1} = {}^nC_r a^{n-r} \cdot x^{n-r} \cdot \frac{1}{x^r} \)
\( \Rightarrow T_{r+1} = {}^nC_r a^{n-r} \cdot x^{n-2r} \)
For \(T_4\), put \(r = 3\)
\( \Rightarrow T_4 = {}^nC_3 a^{n-3} \cdot x^{n-6} \)
Given that, \( T_4 = \frac{5}{2} \)
\( \therefore {}^nC_3 a^{n-3} \cdot x^{n-6} = \frac{5}{2} \) ............ (1)
Clearly R.H.S, of above equation is independent of \(x\)
\( \dots \) put \(n - 6 = 0 \Rightarrow n = 6\)
put \(n = 6\) in equation (1)
\( \Rightarrow {}^6C_3 a^3 \cdot x^0 = \frac{5}{2} \)
\( \Rightarrow \frac{6 \times 5 \times 4}{6} \cdot a^3 = \frac{5}{2} \)
\( \Rightarrow a^3 = \frac{5}{40} = \frac{1}{8} = \frac{1}{2^3} \)
\( \Rightarrow a^3 = \left(\frac{1}{2}\right)^3 \Rightarrow a = \frac{1}{2} \)
\( \dots n = 6 \) & \( a = \frac{1}{2} \)

Question. If \(a_1, a_2, a_3\) and \(a_4\) be the coefficient of four consecutive terms in the expansion of \( (1 + x)^n \), then show that \( \frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2a_2}{a_2 + a_3} \)?
Answer: Expansion \((1 + x)^n\)
General term: \( T_{r+1} = {}^nC_r (1)^{n-r}(x)^r \)
\( T_{r+1} = {}^nC_r x^r \)
Let the four consecutive terms are \(r^{\text{th}}, (r + 1)^{\text{th}}, (r + 2)^{\text{th}}\) and \((r + 3)^{\text{th}}\)
\( T_r = {}^nC_{r-1} \cdot x^{r-1} \Rightarrow \) coefficient of \( {}^nC_{r-1} = a_1 \)
\( T_{r+1} = {}^nC_r \cdot x^r \Rightarrow \) coefficient of \( {}^nC_r = a_2 \)
\( T_{r+2} = {}^nC_{r+1} \cdot x^{r+1} \Rightarrow \) coefficient of \( {}^nC_{r+1} = a_3 \)
\( T_{r+3} = {}^nC_{r+2} \cdot x^{r+2} \Rightarrow \) coefficient of \( {}^nC_{r+2} = a_4 \)
Now, \( a_1 + a_2 = {}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r \)
\( a_2 + a_3 = {}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1} \)
\( a_3 + a_4 = {}^nC_{r+1} + {}^nC_{r+2} = {}^{n+1}C_{r+2} \)
\( {}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r \) (property 1)
taking L.H.S.
\( \frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} \)
\( = \frac{{}^nC_{r-1}}{{}^{n+1}C_r} + \frac{{}^nC_{r+1}}{{}^{n+1}C_{r+2}} \)
\( = \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{(n+1)!}{r!(n+1-r)!}} + \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{(n+1)!}{(r+2)!(n+1-r-2)!}} \)
\( = \frac{n!r!}{(n+1)!(r-1)!} + \frac{n!(r+2)!}{(n+1)!(r+1)!} \)
\( = \frac{n!r(r-1)!}{(n+1)n!(r-1)!} + \frac{n!(r+2)(r+1)!}{(n+1)n!(r+1)!} \)
\( = \frac{r}{n+1} + \frac{r+2}{n+1} \)
\( = \frac{2r+2}{n+1} = \frac{2(r+1)}{n+1} \) ............. (1)
Taking R.H.S. \( \frac{2a_2}{a_2 + a_3} \)
Do yourself and get R.H.S = \( \frac{2(r+1)}{n+1} \) ............. (2)
From eq. (1) and eq. (2), L.H.S. = R.H.S (proved)