Binomial Theorem JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Find the coefficients
(i) \(x^7\) in \(\left(ax^2 + \frac{1}{bx}\right)^{11}\)
(ii) \(x^{-7}\) in \(\left(ax - \frac{1}{bx^2}\right)^{11}\)
(iii) Find the relation between a and b, so that these coefficients are equal.
Answer: (i) \(T_{r+1} = {}^{11}C_r (ax^2)^{11-r} (bx)^{-r} = {}^{11}C_r a^{11-r} b^{-r} x^{22-3r}\)
For \(x^7 \Rightarrow 22 - 3r = 7 \Rightarrow 3r = 15 \Rightarrow r = 5\)
\(\therefore T_6 = T_{5+1} = {}^{11}C_5 a^6 b^{-5} x^7\)
coeff. of \(x^7 = {}^{11}C_5 \frac{a^6}{b^5}\)
(ii) \(T_{r+1} = {}^{11}C_r (ax)^{11-r} (-1)^r (bx^2)^{-r}\)
\(= {}^{11}C_r a^{11-r} b^{-r} (-1)^r x^{11-3r}\)
For \(x^{-7} \Rightarrow 11 - 3r = -7 \Rightarrow 18 = 3r \Rightarrow r = 6\)
\(\therefore T_7 = T_{6+1} = {}^{11}C_6 \frac{a^5}{b^6} x^{-7}\)
coeff. of \(x^{-7} = {}^{11}C_6 \frac{a^5}{b^6}\)
(iii) \({}^{11}C_5 \frac{a^6}{b^5} = {}^{11}C_6 \frac{a^5}{b^6} \Rightarrow a = \frac{1}{b} \Rightarrow ab = 1\)

Question. If the coefficients of \((2r + 4)^{th}\), \((r - 2)^{th}\) terms in the expansion of \((1 + x)^{18}\) are equal, find r.
Answer: In \((1 + x)^{18}\), coeff \(T_{2r+4} = \text{coef } T_{r-2}\)
\(\Rightarrow {}^{18}C_{2r+3} = {}^{18}C_{r-3}\)
\(\Rightarrow 2r + 3 = r - 3 \quad \text{or} \quad 2r + 3 + r - 3 = 18\)
\(\Rightarrow r = -6 \quad \text{or} \quad 3r = 18\)
\(\Rightarrow \text{doesn't exist} \quad \text{or} \quad r = 6\)

Question. If the coefficients of the \(r^{th}\), \((r + 1)^{th}\) and \((r + 2)^{th}\) terms in the expansion of \((1 + x)^{14}\) are in A.P., find r.
Answer: In \((1 + x)^{14}\), coeff. \(r^{th}\), \((r+1)^{th}\), \((r+2)^{th}\) in A.P.
\(\Rightarrow {}^{14}C_{r-1}\), \({}^{14}C_r\), \({}^{14}C_{r+1}\) in A.P.
\(\Rightarrow 2 \cdot {}^{14}C_r = {}^{14}C_{r-1} + {}^{14}C_{r+1}\)
\(\Rightarrow 2 \frac{14!}{r!(14-r)!} = \frac{14!}{(r-1)!(15-r)!} + \frac{14!}{(r+1)!(13-r)!}\)
\(\Rightarrow \frac{2}{r(14-r)} = \frac{1}{(15-r)(14-r)} + \frac{1}{(r+1)r}\)
\(\Rightarrow \frac{2}{r(14-r)} = \frac{r(r+1) + (15-r)(14-r)}{r(r+1)(14-r)(15-r)}\)
\(\Rightarrow 2(r+1)(15-r) = r(r+1) + (15-r)(14-r)\)
\(\Rightarrow 2(15 + 14r - r^2) = r^2 + r + 210 - 29r + r^2\)
\(\Rightarrow 4r^2 - 56r + 180 = 0 \Rightarrow r^2 - 14r + 45 = 0\)
\(\Rightarrow (r-5)(r-9) = 0 \Rightarrow r = 5 \text{ or } 9\)

Question. Find the term independent of x in the expansion of
(a) \(\left[ \sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{2x^2} \right]^{10}\)
(b) \(\left[ \frac{1}{2}x^{1/3} + x^{-1/5} \right]^8\)
Answer: (a) \(T_{r+1} = {}^{10}C_r \left(\frac{x}{3}\right)^{\frac{10-r}{2}} \left(\frac{3^{1/2}}{2x^2}\right)^r = {}^{10}C_r 3^{r-5} 2^{-r} x^{\frac{10-5r}{2}}\)
For constant term \(\Rightarrow \frac{10-5r}{2} = 0 \Rightarrow r = 2\)
\(\therefore T_3 = {}^{10}C_2 \frac{1}{3^3 2^2} = \frac{10 \times 9}{2 \cdot 3^3 \cdot 2^2} = \frac{5}{12}\)
(b) \(T_{r+1} = {}^8C_r \left(\frac{x^{1/3}}{2}\right)^{8-r} (x^{-1/5})^r = {}^8C_r \frac{1}{2^{8-r}} x^{\frac{40-8r}{15}}\)
For constant term \(\Rightarrow \frac{40-8r}{15} = 0 \Rightarrow r = 5\)
\(\therefore T_6 = {}^8C_5 \frac{1}{2^3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2} \frac{1}{2^3} = 7\)

Question. Find the sum of the series
\(\sum_{r=0}^{n} (-1)^r {}^nC_r \left[ \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots \text{up to m terms} \right]\)
Answer: \(\sum_{r=0}^{n} (-1)^r {}^nC_r \left[ \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots \right]\)
\(= \sum_{r=0}^{n} (-1)^r {}^nC_r \left[ \left(1 - \frac{1}{2}\right)^r + \left(1 - \frac{1}{2^2}\right)^r + \left(1 - \frac{1}{2^3}\right)^r + \dots + \left(1 - \frac{1}{2^m}\right)^r \right]\)
\(= \sum_{r=0}^{n} (-1)^r {}^nC_r \left(1 - \frac{1}{2}\right)^r + \sum_{r=0}^{n} (-1)^r {}^nC_r \left(1 - \frac{1}{2^2}\right)^r + \dots + \sum_{r=0}^{n} (-1)^r {}^nC_r \left(1 - \frac{1}{2^m}\right)^r\)
\(= \left[ 1 - \left(1 - \frac{1}{2}\right) \right]^n + \left[ 1 - \left(1 - \frac{1}{2^2}\right) \right]^n + \dots + \left[ 1 - \left(1 - \frac{1}{2^m}\right) \right]^n\)
\(= \left( \frac{1}{2} \right)^n + \left( \frac{1}{2^2} \right)^n + \left( \frac{1}{2^3} \right)^n + \dots + \left( \frac{1}{2^m} \right)^n\)
\(= \frac{1}{2^n} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \dots + \frac{1}{2^{mn}}\)
\(= \frac{\frac{1}{2^n} \left[ 1 - \left(\frac{1}{2^n}\right)^m \right]}{1 - \frac{1}{2^n}} = \frac{1}{2^n} \left[ \frac{1 - \frac{1}{2^{nm}}}{\frac{2^n - 1}{2^n}} \right]\)
\(= \frac{1}{2^n} \frac{(2^{mn} - 1)}{2^{mn}} \frac{2^n}{(2^n - 1)} = \frac{2^{mn} - 1}{(2^n - 1)2^{mn}}\)

Question. If the coefficients of \(2^{nd}\), \(3^{rd}\) and \(4^{th}\) terms in the expansion of \((1 + x)^{2n}\) are in AP, show that \(2n^2 - 9n + 7 = 0\).
Answer: In \((1 + x)^{2n}\), coeff of \(T_2, T_3, T_4\) in AP.
i.e. \({}^{2n}C_1\), \({}^{2n}C_2\), \({}^{2n}C_3\) in AP
\(\Rightarrow 2 \cdot {}^{2n}C_2 = {}^{2n}C_1 + {}^{2n}C_3\)
\(\Rightarrow 2 \frac{(2n)!}{2!(2n-2)!} = \frac{(2n)!}{1!(2n-1)!} + \frac{(2n)!}{3!(2n-3)!}\)
\(\Rightarrow \frac{1}{(2n-2)} = \frac{1}{(2n-1)(2n-2)} + \frac{1}{6}\)
\(\Rightarrow \frac{2n-2}{(2n-1)(2n-2)} = \frac{1}{6} \Rightarrow 6(n-1) = (2n-1)(n-1)\)
\(\Rightarrow (2n-7)(n-1) = 0 \Rightarrow 2n^2 - 9n + 7 = 0\)

Question. Given that \((1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}\), find the values of
(i) \(a_0 + a_1 + a_2 + \dots + a_{2n}\) ;
(ii) \(a_0 - a_1 + a_2 - a_3 \dots + a_{2n}\) ;
(iii) \(a_0^2 - a_1^2 + a_2^2 - a_3^2 + \dots + a_{2n}^2\)
Answer: \((1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}\)
(i) Put x = 1
\(3^n = a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}\)
(ii) Put x = -1
\((1 - 1 + 1)^n = a_0 - a_1 + a_2 + \dots + a_{2n}\)
\(\Rightarrow a_0 - a_1 + a_2 - a_3 + \dots + a_{2n} = 1\)
(iii) Replace \(x \rightarrow -\frac{1}{x}\) in (1)
\(\left(1 - \frac{1}{x} + \frac{1}{x^2}\right)^n = a_0 - a_1\frac{1}{x} + a_2\frac{1}{x^2} - a_3\frac{1}{x^3} + \dots + a_{2n}\frac{1}{x^{2n}}\)
\(\frac{(x^2 - x + 1)^n}{x^{2n}} = a_0 - \frac{a_1}{x} + \frac{a_2}{x^2} - \dots + \frac{a_{2n}}{x^{2n}}\)
Multiply equation (1) & (2) and compare coeff. of \(x^0\)
\(a_0^2 - a_1^2 + a_2^2 - a_3^2 + \dots + a_{2n}^2 = \text{coeff. of } x^{2n} \text{ in } \frac{(1+x^2+x)^n(1+x^2-x)^n}{x^{2n}}\)
\(= \text{coeff. of } x^{2n} \text{ in } \frac{[(1+x^2)^2 - x^2]^n}{x^{2n}}\)
\(= \text{coeff. of } x^{2n} \text{ in } [1+x^2+x^4]^n\)
\(= \text{coeff. of } x^{2n} \text{ in } \sum_{p=0}^{2n} a_p x^{2p} = \text{coeff. of } x^{2n} \text{ is } a_n\)
\(\Rightarrow a_0^2 - a_1^2 + a_2^2 - a_3^2 + \dots + a_{2n}^2 = a_n\)

Question. If a, b, c and d are the coefficients of any four consecutive terms in the expansion of \((1 + x)^n, n \in N\), prove that \(\frac{a}{a+b} + \frac{c}{c+d} = \frac{2b}{b+c}\).
Answer: Let coeff of \((r-1)^{th}, r^{th}, (r+1)^{th}, (r+2)^{th}\) are given a, b, c, d
\({}^nC_{r-2} = a, {}^nC_{r-1} = b, {}^nC_r = c, {}^nC_{r+1} = d\)
then \(\frac{a}{a+b} + \frac{c}{c+d} = \frac{2b}{b+c}\)
\(a+b = {}^nC_{r-2} + {}^nC_{r-1} = {}^{n+1}C_{r-1}\)
\(b+c = {}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r\)
\(c+d = {}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}\)
L.H.S. \(= \frac{{}^nC_{r-2}}{{}^{n+1}C_{r-1}} + \frac{{}^nC_r}{{}^{n+1}C_{r+1}}\)
\(= \frac{r-1}{n+1} + \frac{r+1}{n+1} = \left(\frac{2r}{n+1}\right)\)
R.H.S. \(= \frac{2 \cdot {}^nC_{r-1}}{{}^{n+1}C_r}\)
\(= 2 \cdot \frac{n!}{(r-1)!(n-r+1)!} \cdot \frac{r!(n-r+1)!}{(n+1)!} = 2 \left(\frac{r}{n+1}\right)\)
\(\Rightarrow \text{L.H.S. = R.H.S.}\)

Question. Find the value of x for which the fourth term in the expansion \(\left(5^{\frac{2}{5}\log_5 \sqrt{4^x + 44}} + \frac{1}{5^{\log_5 \sqrt[3]{2^{x-1} + 7}}}\right)^8\) is 336.
Answer: \(\left(5^{\frac{2}{5}\log_5 \sqrt{4^x + 44}} + 5^{-\log_5 \sqrt[3]{2^{x-1} + 7}}\right)^8\)
\(= \left( \left(\sqrt{4^x + 44}\right)^{2/5} + \frac{1}{\sqrt[3]{2^{x-1} + 7}} \right)^8\)
\(= \left( (4^x + 44)^{1/5} + (2^{x-1} + 7)^{-1/3} \right)^8\)
\(T_4 = {}^8C_3 (4^x + 44)^{\frac{8-3}{5}} (2^{x-1} + 7)^{-\frac{3}{3}} = 336\)
\(\Rightarrow {}^8C_3 (4^x + 44)^1 (2^{x-1} + 7)^{-1} = 336\)
\(\Rightarrow 56 \cdot \frac{(4^x + 44)}{(2^{x-1} + 7)} = 336 \Rightarrow 4^x + 44 = 6(2^{x-1} + 7)\)
\(\Rightarrow 2(4^x + 44) = 6(2^x + 14)\)
\(\Rightarrow (2^x)^2 + 44 = 3(2^x) + 42 \Rightarrow (2^x)^2 - 3(2^x) + 2 = 0\)
\(\Rightarrow (2^x - 2)(2^x - 1) = 0 \Rightarrow 2^x = 2 \text{ or } 2^x = 1\)
\(\Rightarrow x = 1 \text{ or } x = 0 \Rightarrow x = 0 \text{ or } 1\)

Question. Prove that : \({}^{n-1}C_r + {}^{n-2}C_r + {}^{n-3}C_r + \dots + {}^rC_r = {}^nC_{r+1}\)
Answer: R.H.S. \(= {}^nC_{r+1} = {}^{n-1}C_r + {}^{n-1}C_{r+1}\)
\(= {}^{n-1}C_r + {}^{n-2}C_r + {}^{n-2}C_{r+1}\)
\(= {}^{n-1}C_r + {}^{n-2}C_r + {}^{n-3}C_r + {}^{n-3}C_{r+1}\)
\(= {}^{n-1}C_r + {}^{n-2}C_r + {}^{n-3}C_r + \dots + {}^{r+1}C_{r+1}\)
\(= {}^{n-1}C_r + {}^{n-2}C_r + {}^{n-3}C_r + \dots + {}^{r+1}C_r + {}^rC_r = \text{L.H.S.}\)

Question. (a) Which is larger : \((99^{50} + 100^{50})\) or \((101)^{50}\).
(b) Show that \({}^{2n-2}C_{n-2} + 2 \cdot {}^{2n-2}C_{n-1} + {}^{2n-2}C_n > \frac{4n}{n+1}\), \(n \in N, n > 2\).
Answer: (a) we will calculate \(101^{50} - 99^{50} = (100 + 1)^{50} - (100 - 1)^{50}\)
\(= (100^{50} + {}^{50}C_1 \cdot 100^{49} + {}^{50}C_2 \cdot 100^{48} + \dots + 1) - (100^{50} - {}^{50}C_1 \cdot 100^{49} + {}^{50}C_2 \cdot 100^{48} - \dots + 1)\)
\(\Rightarrow 101^{50} - 99^{50} = 2 \left[ {}^{50}C_1 \cdot 100^{49} + {}^{50}C_3 \cdot 100^{47} + \dots + {}^{50}C_{49} \cdot 100 \right]\)
\(\Rightarrow 101^{50} - 99^{50} = 100^{50} + 2 \cdot {}^{50}C_3 \cdot 100^{47} + \dots + 2 \cdot {}^{50}C_{49} \cdot 100\)
\(\Rightarrow 101^{50} - 99^{50} = 100^{50} + \text{Positive integer}\)
\(\Rightarrow 101^{50} = 99^{50} + 100^{50} + \text{Positive Integer}\)
\(\Rightarrow 101^{50} > 99^{50} + 100^{50}\)
(b) L.H.S. \(= {}^{2n-2}C_{n-2} + {}^{2n-2}C_{n-1} + {}^{2n-2}C_{n-1} + {}^{2n-2}C_n\)
\(= {}^{2n-1}C_{n-1} + {}^{2n-1}C_n = {}^{2n}C_n > \frac{4n}{n+1}\)

Question. In the expansion of \(\left(1 + x + \frac{7}{x}\right)^{11}\) find the term not containing x.
Answer: Term not containing x:
\(\frac{7}{x}\) | x | 1
0 | 0 | 11 \(\Rightarrow {}^{11}C_0 \cdot 7^0 \cdot {}^{11}C_0 \cdot 1^{11} = 1\)
1 | 1 | 9 \(\Rightarrow {}^{11}C_1 \cdot 7^1 \cdot {}^{10}C_1 \cdot 1^9 = {}^{11}C_2 \cdot {}^2C_1 \cdot 7^1\)
2 | 2 | 7 \(\Rightarrow {}^{11}C_2 \cdot 7^2 \cdot {}^9C_2 \cdot 1 = {}^{11}C_4 \cdot {}^4C_2 \cdot 7^2\)
3 | 3 | 5 \(\Rightarrow {}^{11}C_3 \cdot 7^3 \cdot {}^8C_3 = {}^{11}C_6 \cdot {}^6C_3 \cdot 7^3\)
4 | 4 | 3 \(\Rightarrow {}^{11}C_4 \cdot 7^4 \cdot {}^7C_3 = {}^{11}C_8 \cdot {}^8C_4 \cdot 7^4\)
5 | 5 | 1 \(\Rightarrow {}^{11}C_5 \cdot 7^5 \cdot {}^6C_5 = {}^{11}C_{10} \cdot {}^{10}C_5 \cdot 7^5\)
\(= 1 + \sum_{k=1}^{5} {}^{11}C_{2k} \cdot {}^{2k}C_k 7^k\)

Question. Show that coefficient of \(x^5\) in the expansion of \((1 + x^2)^5 \cdot (1 + x)^4\) is 60.
Answer: For coeff. of \(x^5\)
degrees in of x | \((1 + x^2)^5\) | \((1 + x)^4\)
5 = | 0 | + | Not Possible
5 = | 2 | + | 3
5 = | 4 | + | 1
\(\Rightarrow\) coeff of \(x^2\) in \((1+x^2)^5\) & coeff. of \(x^3\) in \((1+x)^4\) or coeff of \(x^4\) in \((1+x^2)^5\) & coeff of x in \((1+x)^4\)
\(= {}^5C_1 (x^2)^1 \cdot {}^4C_3 x^3 + {}^5C_2 (x^2)^2 \cdot {}^4C_1 x^1\)
\(= (5 \cdot 4 + 10 \cdot 4) x^5\)
coeff. of \(x^5 = (20 + 40) = 60\)

Question. Find the coefficient of \(x^4\) in the expansion of
(i) \((1 + x + x^2 + x^3)^{11}\)
(ii) \((2 - x + 3x^2)^6\)
Answer: (i) \((1 + x + x^2 + x^3)^{11} = [(1 + x) + x^2(1 + x)]^{11} = (1 + x)^{11}(1 + x^2)^{11}\)
For coeff. of \(x^4\)
\((1 + x)^{11}\) | \((1 + x^2)^{11}\)
4 | 0
2 | 2
0 | 4
\(= x^4 \text{ in } (1 + x)^{11} + \{x^2 \text{ in } (1 + x)^{11}\} \{x^2 \text{ in } (1 + x^2)^{11}\} + x^4 \text{ in } (1 + x^2)^{11}\)
\(= {}^{11}C_4 + {}^{11}C_2 \cdot {}^{11}C_1 + {}^{11}C_2 = {}^{11}C_4 + {}^{11}C_2 \cdot 12\)
\(= \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} + \frac{12 \cdot 11 \cdot 10}{2} = 330 + 660 = 990\)
(ii) \({}^6C_0(2 - x)^6 + {}^6C_1(2 - x)^5(3x^2)^1 + {}^6C_2(2 - x)^4(3x^2)^2 + \text{above 4 degrees term of } x^4\)
\(= {}^6C_0 \cdot {}^6C_4 2^2 \cdot (-x)^4 + {}^6C_1 \cdot {}^5C_2 2^3 (-x)^2 3 \cdot x^2 + {}^6C_2 \cdot {}^4C_0 2^4 \cdot 9 \cdot x^4\)
\(= (15 \cdot 4 + 6 \cdot 10 \cdot 24 + 15 \cdot 16 \cdot 9) x^4 = (60 + 1440 + 2160) x^4 = 3660 x^4\)
\(\therefore \text{coeff. of } x^4 \text{ is } 3660\)

Question. Find numerically the greatest term in the expansion of
(i) \((2 + 3x)^9\) when \(x = \frac{3}{2}\)
(ii) \((3 - 5x)^{15}\) when \(x = \frac{1}{5}\)
Answer: (i) \((2 + 3x)^9\) when \(x = \frac{3}{2}\)
\(\Rightarrow \frac{9+1}{|\frac{2}{3x}| + 1} - 1 \leq r \leq \frac{9+1}{|\frac{2}{3x}| + 1}\)
\(\Rightarrow \frac{10}{\frac{4}{9} + 1} - 1 \leq r \leq \frac{10}{\frac{4}{9} + 1} \Rightarrow \frac{77}{13} \leq r \leq \frac{90}{13}\)
\(\Rightarrow 5.9 \leq r \leq 6.9 \Rightarrow r = 6\) (\(\because r \in N\))
\(\therefore T_7 = T_{6+1} = {}^9C_6 (2)^3 \left(3 \cdot \frac{3}{2}\right)^6 = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2} \cdot \frac{2^3 \cdot 3^{12}}{2^6} = \frac{3^{13} \cdot 7}{2}\)
(ii) \((3 - 5x)^{15}\) when \(x = \frac{1}{5}\)
\(\Rightarrow \frac{16}{|\frac{3}{-5x}| + 1} - 1 \leq r \leq \frac{16}{|\frac{3}{-5x}| + 1}\)
\(\Rightarrow 3 \leq r \leq 4 \Rightarrow r = 3 \text{ or } r = 4\)
\(\therefore T_4 = {}^{15}C_3 3^{12} \left( -5 \cdot \frac{1}{5} \right)^3 = -{}^{15}C_3 3^{12} = - \frac{15 \cdot 14 \cdot 13}{3 \cdot 2 \cdot 1} 3^{12} = -455 \times 3^{12}\)
\(T_5 = {}^{15}C_4 3^{11} (-1)^4 = {}^{15}C_4 3^{11} = \frac{15 \cdot 14 \cdot 13 \cdot 12}{3 \cdot 2} = 455 \times 3^{12}\)
\(\therefore |T_4| = |T_5| = 455 \times 3^{12}\)

Question. Given \(s_n = 1 + q + q^2 + \dots + q^n\) and
\(S_n = 1 + \frac{q+1}{2} + \left(\frac{q+1}{2}\right)^2 + \dots + \left(\frac{q+1}{2}\right)^n\), \(q \neq 1\).
Prove that \({}^{n+1}C_1 + {}^{n+1}C_2 \cdot s_1 + {}^{n+1}C_3 \cdot s_2 + \dots + {}^{n+1}C_{n+1} \cdot s_n = 2^n \cdot S_n\).
Answer: \(s_n = \frac{1 - q^{n+1}}{1 - q} \dots (i)\)
\(S_n = \frac{1 - \left(\frac{q+1}{2}\right)^{n+1}}{1 - \left(\frac{q+1}{2}\right)} = \frac{\left[2^{n+1} - (q+1)^{n+1}\right]}{2^n(1-q)} \dots (ii)\)
now L.H.S. \(= \sum_{r=0}^{n} T_{r+1} = \sum_{r=0}^{n} {}^{n+1}C_{r+1} s_r\)
\(= \sum_{r=0}^{n} {}^{n+1}C_{r+1} \left( \frac{1 - q^{r+1}}{1 - q} \right)\)
\(= \frac{1}{1-q} \left[ \sum_{r=0}^{n} {}^{n+1}C_{r+1} - \sum_{r=0}^{n} {}^{n+1}C_{r+1} q^{r+1} \right]\)
\(= \frac{1}{1-q} \left[ \left({}^{n+1}C_1 + {}^{n+1}C_2 + \dots + {}^{n+1}C_{n+1}\right) - \left({}^{n+1}C_1 q^1 + {}^{n+1}C_2 q^2 + \dots + {}^{n+1}C_{n+1} q^{n+1}\right) \right]\)
\(= \frac{1}{1-q} \left[ \left(2^{n+1} - {}^{n+1}C_0\right) - \left((1+q)^{n+1} - {}^{n+1}C_0\right) \right]\)
\(= \frac{1}{1-q} \left[ 2^{n+1} - (1+q)^{n+1} \right] = \frac{1}{1-q} 2^n (1-q) S_n = 2^n S_n = \text{R.H.S.}\)

Question. Prove that the ratio of the coefficient of \(x^{10}\) in \((1 - x^2)^{10}\) & the term independent of x in \(\left(x - \frac{2}{x}\right)^{10}\) is 1 : 32.
Answer: \(T_{r+1} = {}^{10}C_r (-x^2)^r\), \(T_{r+1} = {}^{10}C_r (x)^{10-r} \left(-\frac{2}{x}\right)^r\)
For \(x^{10} \Rightarrow 2r = 10 \Rightarrow r = 5\)
\(T_6 = ({}^{10}C_5) \cdot x^{10} (-1)^r = {}^{10}C_r (-2)^r x^{10-2r}\)
For constant term \(10 - 2r = 0 \Rightarrow r = 5\)
\(\therefore\) coeff. of \(x^{10} = -{}^{10}C_5\)
\(T_6 = {}^{10}C_5 (-2)^5 x^0\)
\(\therefore\) coeff. of constant term \(= -{}^{10}C_5 \cdot 2^5\)
required ratio \(= \frac{-{}^{10}C_5}{-{}^{10}C_5 \cdot 2^5} = \frac{1}{2^5} = \frac{1}{32} = 1:32\)

Advanced Subjective Questions

Question. \( C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \dots + (-1)^n \frac{C_n}{n + 1} = \frac{1}{n + 1} \)
Answer: L.H.S. \( = C_0 - \frac{C_1}{2} + \frac{C_2}{3} - \dots + (-1)^n \frac{C_n}{n + 1} \)
\( \sum T_{p+1} = \sum_{0}^n \frac{(-1)^P \cdot {}^nC_p}{p + 1} = \frac{1}{(n + 1)} \sum_{0}^n {}^{n+1}C_{p+1} (-1)^p \)
\( = \frac{1}{(n + 1)} [{}^{n+1}C_1 - {}^{n+1}C_2 + {}^{n+1}C_3 - {}^{n+1}C_4 \dots + (-1)^n \cdot {}^{n+1}C_{n+1}] \)
\( = \frac{1}{(n + 1)} [{}^{n+1}C_0 - ({}^{n+1}C_0 - {}^{n+1}C_1 + {}^{n+1}C_2 + \dots - (-1)^n \cdot {}^{n+1}C_{n+1})] \)
\( = \frac{1}{(n + 1)} [1 - 0] = \frac{1}{(n + 1)} = \text{R.H.S.} \)

Question. \( C_0 - 2C_1 + 3C_2 - 4C_3 + \dots + (-1)^n (n + 1) C_n = 0 \)
Answer: L.H.S. \( = \sum_{P=0}^n T_{P+1} = \sum_{P=0}^n (-1)^P (P + 1) C_P = \sum_{P=0}^n (-1)^P P \cdot C_P + \sum_{P=0}^n (-1)^P \cdot C_P \)
\( = \sum_{P=0}^n (-1)^p p \cdot \frac{n}{p} {}^{n-1}C_{p-1} + \sum_{P=0}^n {}^nC_P (-1)^p \)
\( = (-n) \sum_{P=1}^n {}^{n-1}C_{p-1} (-1)^{p-1} + \sum_{P=0}^n {}^nC_p (-1)^p = (-n) [1 - 1]^{n-1} + [1 - 1]^n = 0 + 0 = 0 = \text{R.H.S.} \)

Question. \( (n - 1)^2 \cdot C_1 + (n - 3)^2 \cdot C_3 + (n - 5)^2 \cdot C_5 + \dots = n(n + 1) 2^{n - 3} \)
Answer: L.H.S. \( = \sum_{r=1}^n (n-r)^2 \cdot C_r = \sum_{r=1}^n (n^2 - 2nr + r^2) C_r = n^2 \sum_{1}^n C_r - 2n \sum_{1}^n r \cdot C_r + \sum_{1}^n r^2 \cdot C_r \)
\( = n^2 (2^n - 1) + (1 - 2n) n \cdot 2^{n-1} + n(n - 1) 2^{n-2} = n^2 2^n - n^2 + n \cdot 2^{n-1} - n^2 2^n + n(n - 1)2^{n-2} = -n^2 + n(n + 1) 2^{n-2} \dots(1) \)
Again L.H.S. \( = \sum_{r=1}^n (n-r)^2 C_r (-1)^{r+1} = \sum_{r=1}^n (n^2 - 2nr + r^2) C_r (-1)^{r+1} = n^2 \sum_{r=1}^n {}^nC_r (-1)^{r+1} + (1 - 2n) \sum_{1}^n r \cdot C_r (-1)^{r+1} + n (n - 1) \sum_{1}^n {}^{n-2}C_{r-2} (-1)^{r+1} = -n^2 [(1 - 1)^n - 1] + (1 - 2n)n [1 - 1]^{n-1} - n (n - 1) [1 - 1]^{n-2} = -n^2 [0 - 1] + 0 + 0 = n^2 \dots(2) \)
Adding (1) & (2): \( 2[(n - 1)^2 \cdot C_1 + (n - 3)^2 \cdot C_3 + (n - 5)^2 C_5 + \dots] = -n^2 + n (n + 1) 2^{n-2} + n^2 \)
\( \Rightarrow (n - 1)^2 \cdot C_1 + (n - 3)^2 \cdot C_3 + (n - 5)^2 C_5 + \dots = \frac{n(n+1)2^{n-2}}{2} = n (n + 1) 2^{n-3} \)

Question. \( 1 \cdot C_0^2 + 3 \cdot C_1^2 + 5 \cdot C_2^2 + \dots + (2n + 1) C_n^2 = \frac{(n + 1)(2n)!}{n! n!} \)
Answer: Let \( S = 1 \cdot C_0^2 + 3 \cdot C_1^2 + 5 \cdot C_2^2 + \dots + (2n + 1) \cdot C_n^2 \)
\( S = (2n + 1) C_0^2 + (2n - 1) C_1^2 + (2n - 2) C_2^2 + \dots + 1 \cdot C_n^2 \)
Adding: \( 2S = (2n + 2) [C_0^2 + C_1^2 + C_2^2 + \dots + C_n^2] \)
\( 2S = 2(n + 1) \cdot {}^{2n}C_n \)
\( S = (n + 1) \cdot {}^{2n}C_n = \frac{(n + 1)(2n)!}{n! \cdot n!} = \text{R.H.S.} \)

Question. If \( a_0, a_1, a_2, \dots \) be the coefficients in the expansion of \( (1 + x + x^2)^n \) in ascending powers of \( x \),
then prove that: \( a_0 a_1 - a_1 a_2 + a_2 a_3 - \dots = 0 \)
Answer: \( (1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_nx^n + \dots + a_{2n}x^{2n} \dots(i) \)
Replace \( x \rightarrow \left( -\frac{1}{x} \right) \)
\( \left( 1 - \frac{1}{x} + \frac{1}{x^2} \right)^n = a_0 - \frac{a_1}{x} + \frac{a_2}{x^2} - \dots + \frac{a_{2n}}{x^{2n}} \)
\( \Rightarrow \frac{(x^2 - x + 1)^n}{x^{2n}} = \frac{a_0x^{2n} - a_1x^{2n-1} + a_2x^{2n-1} + \dots + a_{2n}}{x^{2n}} \dots(ii) \)
Multiply (i) & (ii) then L.H.S are \( \{(x^2 + 1)^2 - x^2\}^n = (a_0^2 - a_1^2 + a_2^2 - a_3^2 + \dots + a_{2n}^2)x^{2n} + (\dots) \)
\( (1 + x^2 + x^4)^n = (a_0^2 - a_1^2 + a_2^2 - a_3^2 + \dots + a_{2n}^2)x^{2n} + (\dots) \dots(iii) \)
Multiply (i) & (ii) and compare of \( x^1 \) or \( (-x^{-1}) \):
\( a_0a_1 - a_1a_2 + a_2a_3 \dots = x^1 \text{ in } \frac{(1 + x^2 + x^4)^n}{x^{2n}} = x^{2n+1} \text{ in } (1 + x^2 + x^4)^n = 0 \) [\( \because \) all degree of in \( (1 + x^2 + x^4)^n \) is even]

Question. If \( a_0, a_1, a_2, \dots \) be the coefficients in the expansion of \( (1 + x + x^2)^n \) in ascending powers of \( x \),
then prove that: \( a_0 a_2 - a_1 a_3 + a_2 a_4 - \dots + a_{2n - 2} a_{2n} = a_{n + 1} \) or \( a_{n - 1} \)
Answer: Multiply (i) & (ii) and compare of \( x^2 \) or \( x^{-2} \):
\( a_0a_2 - a_1a_3 + a_2 a_4 - \dots + a_{2n-2} a_{2n} = x^2 \text{ in } \frac{(1 + x^2 + x^4)^n}{x^{2n}} = x^{2n+2} \text{ in } (1 + x^2 + x^4)^n = a_{n+1} \)
or \( = x^{-2} \text{ in } \frac{(1 + x^2 + x^4)^n}{x^{2n}} = x^{2n-2} \text{ in } (1 + x^2 + x^4)^n = a_{n-1} \)

Question. If \( a_0, a_1, a_2, \dots \) be the coefficients in the expansion of \( (1 + x + x^2)^n \) in ascending powers of \( x \),
then prove that: \( E_1 = E_2 = E_3 = 3^{n - 1} \) where \( E_1 = a_0 + a_3 + a_6 + \dots \); \( E_2 = a_1 + a_4 + a_7 + \dots \) & \( E_3 = a_2 + a_5 + a_8 + \dots \)
Answer: Put \( x = \omega \) in equation (i)
\( (1 + \omega + \omega^2)^n = a_0 + a_1\omega + a_2\omega^2 + a_3\omega^3 + \dots + a_{2n}\omega^{2n} \)
\( \because \omega \) is cube root of unity \( \therefore \omega^3 = 1, 1 + \omega + \omega^2 = 0, \omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)
\( \Rightarrow 0^n = (a_0 + a_3 + a_6 + \dots)\omega^3 + (a_1 + a_4 + a_7 + \dots) \omega + (a_2 + a_5 + a_8 + \dots)\omega^2 \)
\( \Rightarrow E_1 + E_2\omega + E_3\omega^2 = 0 \)
\( \therefore E_1 - \frac{E_2}{2} - \frac{E_3}{2} = 0 \) & \( E_2 - E_3 = 0 \Rightarrow E_2 = E_3 \)
\( \Rightarrow E_1 = E_2 \therefore E_1 = E_2 = E_3 \)
Put \( x = 1 \) in (i): \( a + a_1 + a_2 + a_3 + \dots a_{2n} = (1 + 1 + 1)^n = 3^n \)
\( \Rightarrow (a_0 + a_3 + a_6 + \dots) + (a_1 + a_4 + a_7 + \dots ) + (a_2 + a_5 + a_8 + \dots) = 3^n \)
\( \Rightarrow E_1 + E_2 + E_3 = 3^n \Rightarrow 3E_1 = 3^n \therefore E_1 = 3^{n-1} = E_2 = E_3 \)

Question. Prove that : \( \sum_{r=0}^{n - 2} ({}^nC_r \cdot {}^nC_{r + 2}) = \frac{(2n)!}{(n - 2)! (n + 2)!} \)
Answer: \( (1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n \dots(i) \)
\( (x + 1)^n = C_0x^n + C_1x^{n-1} + C_2x^{n-2} + \dots + C_n \dots(ii) \)
Multiply (i) & (ii) and compare coeff. of \( x^{n+2} \) or \( x^{n-2} \):
\( C_0C_2 + C_1C_3 + C_2C_4 + \dots + C_{n-2}C_n = x^{n+2} \text{ or } x^{n-2} \text{ in } (1 + x)^{2n} = {}^{2n}C_{n-2} = \frac{(2n)!}{(n - 2)!(n + 2)!} \)

Question. If \( (1 + x)^n = C_0 + C_1 x + C_2x^2 + \dots + C_nx^n \),
then show that the sum of the products of the \( C_i's \) taken two at a time, represented by \( \sum_{0 \le i < j \le n} \sum C_i C_j \) is equal to \( 2^{2n - 1} - \frac{2n!}{2(n!)^2} \).
Answer: \( \sum \sum_{0 \le i < j \le n} C_i C_j = 2^{2n-1} - \frac{(2n)!}{2(n!)^2} \)
\( \sum_{r=1}^n (C_0C_r + C_1C_{r+1} + C_2C_{r+2} + \dots + C_{n-r} \cdot C_n) = \sum_{r=1}^n {}^{2n}C_{n+r} \)
\( \sum_{r=1}^n {}^{2n}C_{n+r} = {}^{2n}C_{n+1} + {}^{2n}C_{n+2} + {}^{2n}C_{n+3} + \dots + {}^{2n}C_{2n} \)
\( = \frac{1}{2} [({}^{2n}C_0 + {}^{2n}C_1 + \dots + {}^{2n}C_{2n}) - {}^{2n}C_n] \)
\( = \frac{1}{2} [2^{2n} - {}^{2n}C_n] = 2^{2n-1} - \frac{(2n)!}{2 \cdot (n!)^2} \)

Question. \( \sqrt{C_1} + \sqrt{C_2} + \sqrt{C_3} + \dots + \sqrt{C_n} \le 2^{n - 1} + \frac{n - 1}{2} \).
Answer: Multiply 2 both side:
\( \Rightarrow 2\sqrt{C_1} + 2\sqrt{C_2} + 2\sqrt{C_3} + \dots + 2\sqrt{C_n} \le 2^n + (n - 1) \)
\( \Rightarrow 2\sqrt{C_1} + 2\sqrt{C_2} + 2\sqrt{C_3} + \dots + \sqrt{C_n} \le C_0 + C_1 + C_2 + C_3 + \dots + C_n + n - 1 \)
\( \{\because 2^n = C_0 + C_1 + C_2 + \dots + C_n \text{ & } C_0 = 1\} \)
\( \Rightarrow 2\sqrt{C_1} + 2\sqrt{C_2} + 2\sqrt{C_3} + \dots + 2\sqrt{C_n} \le C_1 + C_2 + C_3 + \dots C_n + (1 + 1 + 1 + \dots n \text{ times }) \)
\( \Rightarrow 2\sqrt{C_1} + 2\sqrt{C_2} + 2\sqrt{C_3} + \dots + 2\sqrt{C_n} \le (C_1 + 1) + (C_2 + 1) + (C_3 + 1) + \dots + (C_n + 1) \)
\( \because \text{A.M.} \ge \text{G.M.} \Rightarrow \frac{C_1 + 1}{2} \ge \sqrt{C_1 \times 1} \Rightarrow C_1 + 1 \ge 2\sqrt{C_1} \)
\( \therefore 2\sqrt{C_1} + 2\sqrt{C_2} + 2\sqrt{C_3} + \dots + 2\sqrt{C_n} \le (C_1 + 1) + (C_2 + 1) + \dots + (C_n + 1) \quad \text{H.P.} \)

Question. \( \sqrt{C_1} + \sqrt{C_2} + \sqrt{C_3} + \dots + \sqrt{C_n} \le [n(2^n - 1)]^{1/2} \) for \( n \ge 2 \).
Answer: \( \{ \because \text{Root mean square} \ge \text{A.M.} \} \)
\( \Rightarrow \sqrt{\frac{(\sqrt{C_1})^2 + (\sqrt{C_2})^2 + \dots + (\sqrt{C_n})^2}{n}} \ge \frac{\sqrt{C_1} + \sqrt{C_2} + \sqrt{C_3} + \dots + \sqrt{C_n}}{n} \)
\( \Rightarrow \frac{C_1 + C_2 + C_3 + \dots + C_n}{n} \ge \left( \frac{\sqrt{C_1} + \sqrt{C_2} + \dots + \sqrt{C_n}}{n} \right)^2 \)
\( \Rightarrow \sqrt{C_1} + \sqrt{C_2} + \dots + \sqrt{C_n} \le \sqrt{n(C_1 + C_2 + \dots + C_n)} \)
\( \Rightarrow \sqrt{C_1} + \sqrt{C_2} + \dots + \sqrt{C_n} \le \sqrt{n(C_0 + C_1 + C_2 + \dots + C_n - C_0)} \)
\( \Rightarrow \sqrt{C_1} + \sqrt{C_2} + \dots + \sqrt{C_n} \le [n(2^n - 1)]^{1/2} \)

JEE Problems

Question. If in the expansion of \( (1 + x)^m (1 - x)^n \), the co-efficients of \( x \) and \( x^2 \) are 3 and \( -6 \) respectively, then \( m \) is
(a) 6
(b) 9
(c) 12
(d) 24
Answer: (c) 12

Question. For \( 2 \leq r \leq n \), \( \binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} = \)
(a) \( \binom{n+1}{r-1} \)
(b) \( 2\binom{n+1}{r+1} \)
(c) \( 2\binom{n+2}{r} \)
(d) \( \binom{n+2}{r} \)
Answer: (d) \( \binom{n+2}{r} \)

Question. For any positive integers \( m, n \) (with \( n \geq m \)), let \( \binom{n}{m} = {}^nC_m \).
Prove that \( \binom{n}{m} + \binom{n-1}{m} + \binom{n-2}{m} + \dots + \binom{m}{m} = \binom{n+1}{m+1} \).
Hence or otherwise prove that, \( \binom{n}{m} + 2\binom{n-1}{m} + 3\binom{n-2}{m} + \dots + (n - m + 1)\binom{m}{m} = \binom{n+2}{m+2} \)
Answer: LHS = \( \binom{n}{m} + \binom{n-1}{m} + \binom{n-2}{m} + \dots + \binom{m}{m} \). Using the identity \( \binom{r}{r} = \binom{r+1}{r+1} \),
we have \( \binom{m}{m} + \binom{m+1}{m} = \binom{m+2}{m+1} \). Continuing this process of adding subsequent terms, we get \( \binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} = \) R.H.S.
For the second part, let \( S = \binom{n}{m} + 2\binom{n-1}{m} + 3\binom{n-2}{m} + \dots + (n - m + 1)\binom{m}{m} \). This can be written as:
\( S = [\binom{n}{m} + \binom{n-1}{m} + \dots + \binom{m}{m}] + [\binom{n-1}{m} + \binom{n-2}{m} + \dots + \binom{m}{m}] + \dots + \binom{m}{m} \)
Using the result from the first part, \( S = \binom{n+1}{m+1} + \binom{n}{m+1} + \binom{n-1}{m+1} + \dots + \binom{m+1}{m+1} \).
Applying the identity again, \( S = \binom{n+2}{m+2} = \) R.H.S.

Question. Find the largest co-efficient in the expansion of \( (1 + x)^n \), given that the sum of co-efficients of the terms in its expansion is 4096.
Answer: The sum of coefficients is \( 2^n = 4096 \Rightarrow 2^n = 2^{12} \Rightarrow n = 12 \). For \( n = 12 \),
the largest coefficient is the middle term coefficient, which is \( {}^{12}C_{12/2} = {}^{12}C_6 \).

Question. In the binomial expansion of \( (a - b)^n, n \geq 5 \), the sum of the 5th and 6th terms is zero. Then \( \frac{a}{b} \) equals.
(a) \( \frac{n - 5}{6} \)
(b) \( \frac{n - 4}{5} \)
(c) \( \frac{5}{n - 4} \)
(d) \( \frac{6}{n - 5} \)
Answer: (b) \( \frac{n - 4}{5} \)

Question. Find the coefficient of \( x^{49} \) in the polynomial \( (x - \frac{C_1}{C_0})(x - 2^2 \frac{C_2}{C_1})(x - 3^2 \frac{C_3}{C_2}) \dots (x - 50^2 \frac{C_{50}}{C_{49}}) \) where \( C_r = {}^{50}C_r \).
Answer: The coefficient of \( x^{49} \) in a polynomial of degree 50 is the negative sum of the roots. Sum of roots \( = \sum_{r=1}^{50} r^2 \frac{C_r}{C_{r-1}} \).
Since \( \frac{C_r}{C_{r-1}} = \frac{50 - r + 1}{r} \), the sum is \( \sum_{r=1}^{50} r^2 \frac{51 - r}{r} = \sum_{r=1}^{50} (51r - r^2) = 51 \sum_{r=1}^{50} r - \sum_{r=1}^{50} r^2 \).
\( = 51 \frac{50 \times 51}{2} - \frac{50 \times 51 \times 101}{6} = 65025 - 42925 = 22100 \).
Coefficient of \( x^{49} = -22100 \).

Question. The sum \( \sum_{i=0}^m \binom{10}{i}\binom{20}{m-i} \), (where \( \binom{p}{q} = 0 \) if \( p < q \)) is maximum when \( m \) is
(a) 5
(b) 10
(c) 15
(d) 20
Answer: (c) 15

Question. Coefficient of \( t^{24} \) in the expansion of \( (1 + t^2)^{12} (1 + t^{12}) (1 + t^{24}) \) is
(a) \( {}^{12}C_6 + 2 \)
(b) \( {}^{12}C_6 + 1 \)
(c) \( {}^{12}C_6 \)
(d) None of the options
Answer: (a) \( {}^{12}C_6 + 2 \)