Binomial Theorem JEE Mathematics Worksheets Set 02

Subjective Questions

Question. Find the term independent of x in the expansion of \((1 + x + 2x^3) \left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9\).
Answer: Term independent of x in \((1 + x + 2x^3) \left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9\)
\(= 1 \cdot (\text{x}^0 \text{ in B}) + 1 \cdot (\text{x}^{-1} \text{ in B}) + 2 \cdot (\text{x}^{-3} \text{ in B})\)
\(T_{r+1} = {}^9C_r \left(\frac{3}{2}\right)^{9-r} x^{18-2r} \left(-\frac{1}{3}\right)^r x^{-r} = {}^9C_r \frac{3^{9-2r}}{2^{9-r}} (-1)^r x^{18-3r}\)
For constant term \(x^0 \Rightarrow 18 - 3r = 0 \Rightarrow r = 6\)
\(\therefore T_7 = {}^9C_6 \frac{3^{-3}}{2^3} x^0\)
For constant term \(x^{-1} \Rightarrow 18 - 3r = -1 \Rightarrow 19 = 3r\) not possible
For constant term \(x^{-3} \Rightarrow 18 - 3r = -3 \Rightarrow 21 = 3r \Rightarrow r = 7\)
\(\therefore T_8 = {}^9C_7 \frac{3^{-5}}{2^2} (-1)^7 x^{-3}\)
\(= 1 \cdot {}^9C_6 \frac{1}{2^3 \cdot 3^3} - 2 \cdot {}^9C_7 \frac{1}{2^2 \cdot 3^5}\)
\(= \frac{9 \cdot 8 \cdot 7}{3 \cdot 2} \cdot \frac{1}{2^3 \cdot 3^3} - 2 \cdot \frac{9 \cdot 8}{2 \cdot 1} \cdot \frac{1}{2^2 \cdot 3^5}\)
\(= \frac{9 \cdot 8}{2^3 \cdot 3^3} \left( \frac{7}{6} - \frac{2}{9} \right) = \frac{1}{3} \left( \frac{21 - 4}{18} \right) = \frac{17}{54}\)

Question. Let \((1 + x^2)^2 \cdot (1 + x)^n = \sum_{K=0}^{n+4} a_K x^K\). If \(a_1\), \(a_2\) and \(a_3\) are in AP, find n.
Answer: L.H.S. \(= (1 + x^2)^2(1 + x)^n = (1 + 2x^2 + x^4)(1 + x)^n\)
\(= (1 + 2x^2 + x^4)\left(1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots\right)\)
\(1 + nx + (2 + {}^nC_2)x^2 + (2n + {}^nC_3)x^3 + \text{higher terms of x}\)
By comparing with \(a_0 x^0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots\)
\(a_0 = 1, a_1 = n, a_2 = 2 + {}^nC_2, a_3 = 2n + {}^nC_3\)
\(\therefore a_1, a_2, a_3\) in AP \(\Rightarrow 2a_2 = a_1 + a_3\)
\(\Rightarrow 4 + 2 \cdot {}^nC_2 = n + 2n + {}^nC_3\)
\(\Rightarrow 4 + 2 \frac{n(n-1)}{2} = 3n + \frac{n(n-1)(n-2)}{6}\)
\(\Rightarrow 6(n^2 - 4n + 4) = n(n-1)(n-2)\)
\(\Rightarrow (n-2)[n^2 - 7n + 12] = 0\)
\(\Rightarrow (n-2)(n-3)(n-4) = 0 \Rightarrow n = 2 \text{ or } 3 \text{ or } 4\)

Question. If the coefficient of \(a^{r-1}\), \(a^r\), \(a^{r+1}\) in the expansion of \((1 + a)^n\) are in arithmetic progression then prove that \(n^2 - n(4r + 1) + 4r^2 - 2 = 0\).
Answer: Coeff. \(a^{r-1}, a^r, a^{r+1}\) in \((1 + a)^n\) are in A.P.
\(\Rightarrow {}^nC_{r-1}, {}^nC_r, {}^nC_{r+1}\) in A.P.
\(\Rightarrow 2 \cdot {}^nC_r = {}^nC_{r-1} + {}^nC_{r+1}\)
\(\Rightarrow 2 \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r+1)!(r-1)!} + \frac{n!}{(n-r)!(r+1)!}\)
\(\Rightarrow \frac{2}{r(n-r)} = \frac{1}{(n-r+1)(n-r)} + \frac{1}{r(r+1)}\)
\(\Rightarrow \frac{1}{n-1} \left[ \frac{2}{r} - \frac{1}{n-r+1} \right] = \frac{1}{r(r+1)}\)
\(\Rightarrow \frac{2n - 3r + 2}{(n-r)(n-r+1)} = \frac{1}{(r+1)}\)
\(\Rightarrow n^2 - n(4r+1) + 4r^2 - 2 = 0\)

Question. If \({}^nJ_r = \frac{(1-x^n)(1-x^{n-1})(1-x^{n-2})\dots(1-x^{n-r+1})}{(1-x)(1-x^2)(1-x^3)\dots(1-x^r)}\), prove that \({}^nJ_{n-r} = {}^nJ_r\).
Answer: R.H.S. \(= {}^nJ_r = \frac{(1-x^n)(1-x^{n-1})\dots(1-x^{n-r+1})}{(1-x)(1-x^2)\dots(1-x^r)}\)
\(= \frac{(1-x^n)(1-x^{n-1})\dots(1-x^{n-r+1})}{(1-x)(1-x^2)\dots(1-x^r)} \cdot \frac{(1-x^{n-r})\dots(1-x)}{(1-x^{n-r})\dots(1-x)}\)
\(= \frac{(1-x^n)(1-x^{n-1})\dots(1-x)}{(1-x)(1-x^2)\dots(1-x^r) (1-x^{n-r})\dots(1-x)}\)
\(= \frac{(1-x^n)(1-x^{n-1})\dots(1-x)}{(1-x)\dots(1-x^{n-r})(1-x^{n-r+1})\dots(1-x^n)}\)
\(= \frac{(1-x^{n-r})(1-x^{n-r-1})\dots(1-x)}{(1-x^{r+1})\dots(1-x^{n-r})}\)
by given definition \(= {}^nJ_{n-r} = \text{L.H.S.}\)
Aliter: \(\frac{{}^nJ_{n-r}}{{}^nJ_r} = 1 \Rightarrow {}^nJ_{n-r} = {}^nJ_r\)

Question. Prove that \(\sum_{K=0}^{n} {}^nC_K \sin Kx \cdot \cos(n - K)x = 2^{n-1} \sin nx\).
Answer: Let \(S = \sum_{k=0}^{n} {}^nC_k \sin kx \cdot \cos(n-k)x\)
\(S = {}^nC_0 \sin 0x \cos nx + {}^nC_1 \sin 1x \cos (n-1)x + \dots + {}^nC_n \sin nx \cos 0x \quad \dots (1)\)
arrange reverse order
\(S = {}^nC_0 \sin nx \cos 0x + {}^nC_1 \sin (n-1)x \cos 1x + \dots + {}^nC_n \sin 0x \cos nx \quad \dots (2)\)
adding (1) & (2) {and apply \(\sin(A+B)\)}
\(\Rightarrow 2S = {}^nC_0 \sin(nx + 0x) + {}^nC_1 \sin(nx - x + x) + \dots + {}^nC_n \sin(0x + nx)\)
\(\Rightarrow 2S = {}^nC_0 \sin nx + {}^nC_1 \sin nx + {}^nC_2 \sin nx + \dots + {}^nC_n \sin nx\)
\(\Rightarrow 2S = [{}^nC_0 + {}^nC_1 + {}^nC_2 + \dots + {}^nC_n] \sin nx\)
\(\Rightarrow 2S = 2^n \sin nx \Rightarrow S = 2^{n-1} \sin nx = \text{R.H.S.}\)

Question. The expressions \(1 + x, 1 + x + x^2, 1 + x + x^2 + x^3, \dots, 1 + x + x^2 + \dots + x^n\) are multiplied together and the terms of the product thus obtained are arranged in increasing powers of x in the form of \(a_0 + a_1x + a_2x^2 + \dots\), then
(a) how many terms are there in the product
(b) show that the coefficients of the terms in the product, equidistant from the beginning and end are equal.
(c) show that the sum of the odd coefficients = the sum of the even coefficients = \(\frac{(n+1)!}{2}\).
Answer: If \((1+x)(1+x+x^2)\dots(1+x+\dots+x^n) = a_0 + a_1x + a_2x^2 + \dots \dots(1)\)
(a) max power of x in L.H.S. = sum of max degree in each factor
\(= x^1 \cdot x^2 \cdot x^3 \dots x^n = x^{1+2+3+\dots+n} = x^{n(n+1)/2} = x^k\)
\(\text{R.H.S.} = a_0 + a_1x + a_2x^2 + \dots + a_{n(n+1)/2} x^{\frac{n(n+1)}{2}}\)
No. of terms are \(= \frac{n(n+1)}{2} + 1 = \frac{n^2+n+2}{2}\)
(b) Replace \(x \rightarrow \frac{1}{x}\) in equation (1)
\(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x}+\frac{1}{x^2}\right)\dots\left(1+\frac{1}{x}+\dots+\frac{1}{x^n}\right) = a_0 + \frac{a_1}{x} + \frac{a_2}{x^2} + \dots + \frac{a_k}{x^k}\)
\(\frac{(1+x)}{x} \cdot \frac{(1+x+x^2)}{x^2} \dots \frac{(1+x+\dots+x^n)}{x^n} = \frac{a_0 x^k + a_1 x^{k-1} + \dots + a_k}{x^k} \dots(2)\)
By comparing (1) & (2)
\(a_0 = a_k \Rightarrow \text{constant}\)
\(a_1 = a_{k-1} \Rightarrow \text{coeff. of x}\)
\(a_2 = a_{k-2} \Rightarrow \text{coeff. of } x^2\)
(c) put x = 1, by (1)
\(a_0 + a_1 + a_2 + a_3 + \dots = 2 \cdot 3 \cdot 4 \dots (n+1)\)
\(\Rightarrow a_0 + a_1 + a_2 + a_3 + \dots = (n+1)! \dots(3)\)
Put x = -1, by (1)
\(a_0 - a_1 + a_2 - a_3 + \dots = (1-1)(1-1+1)\dots = 0 \dots(4)\)
Add (3) & (4)
\(2[a_0 + a_2 + a_4 + \dots] = (n+1)! \Rightarrow a_0 + a_2 + a_4 + \dots = \frac{(n+1)!}{2} \dots(5)\)
Subtract (3) & (4)
\(2[a_1 + a_3 + a_5 + \dots] = (n+1)! \Rightarrow a_1 + a_3 + a_5 + \dots = \frac{(n+1)!}{2} \dots(6)\)
From (5) & (6) H.P.

Question. Find the coefficients of
(a) \(x^6\) in the expansion of \((ax^2 + bx + c)^9\)
(b) \(x^2 y^3 z^4\) in the expansion of \((ax - by + cz)^9\).
(c) \(a^2 b^3 c^4 d\) in the expansion of \((a - b - c + d)^{10}\)
Answer: (a) \(x^6\) in \((ax^2 + bx + c)^9\)
Powers of \(ax^2\), \(bx\), \(c\):
\(ax^2\) | \(bx\) | \(c\)
0 | 6 | 3 \(\Rightarrow \frac{9!}{0!6!3!} a^0 b^6 c^3 = 84 b^6 c^3\)
1 | 4 | 4 \(\Rightarrow \frac{9!}{1!4!4!} a^1 b^4 c^4 = 630 a b^4 c^4\)
2 | 2 | 5 \(\Rightarrow \frac{9!}{2!2!5!} a^2 b^2 c^5 = 756 a^2 b^2 c^5\)
3 | 0 | 6 \(\Rightarrow \frac{9!}{3!0!6!} a^3 b^0 c^6 = 84 a^3 c^6\)
\(= 84b^6c^3 + 630ab^4c^4 + 756a^2b^2c^5 + 84a^3c^6\)
(b) \(x^2 y^3 z^4\) in \((ax - by + cz)^9\)
\(= \frac{9!}{2!3!4!} a^2 (-b)^3 c^4 = -36 \cdot 35 a^2 b^3 c^4 = -1260 a^2 b^3 c^4\)
(c) \(a^2 b^3 c^4 d\) in \((a - b - c + d)^{10}\)
\(= \frac{10!}{2!3!4!1!} (1)^2 (-1)^3 (-1)^4 (1)^1 = -45 \cdot 56 \cdot 5 \cdot 1 = -2520 \cdot 5 = -12600\)

Question. If \(\sum_{r=0}^{2n} a_r(x-2)^r = \sum_{r=0}^{2n} b_r(x-3)^r\) and \(a_k = 1\) for all \(k \ge n\), then show that \(b_n = {}^{2n+1}C_{n+1}\).
Answer: Let \(x-3 = y \Rightarrow (x-2) = (x-3) + 1 = y+1\)
\(\Rightarrow \sum_{r=0}^{2n} a_r(1+y)^r = \sum_{r=0}^{2n} b_r y^r\)
\(\Rightarrow a_0 + a_1(1+y) + a_2(1+y)^2 + \dots + a_{n-1}(1+y)^{n-1} + a_n(1+y)^n + a_{n+1}(1+y)^{n+1} + \dots + a_{2n}(1+y)^{2n}\)
\(= b_0 + b_1 y + b_2 y^2 + \dots + b_n y^n + \dots + b_{2n} y^{2n}\)
Compare coeff. of \(y^n\) both sides.
In L.H.S. coeff. of \(y^n\)
\({}^nC_n\) in \((1+y)^n\)
\({}^{n+1}C_n\) in \((1+y)^{n+1}\)
\({}^{2n}C_n\) in \((1+y)^{2n}\)
Coeff. of \(y^n\) in L.H.S. = Coeff. of \(y^n\) in R.H.S.
\({}^nC_n + {}^{n+1}C_n + {}^{n+2}C_n + \dots + {}^{2n}C_n = b_n\)
\(\{ {}^nC_n = 1 = {}^{n+1}C_{n+1} \}\)
\(\Rightarrow {}^{n+1}C_{n+1} + {}^{n+1}C_n + {}^{n+2}C_n + \dots + {}^{2n}C_n = b_n\)
\(\Rightarrow {}^{n+2}C_{n+1} + {}^{n+2}C_n + \dots + {}^{2n}C_n = b_n\)
\(\Rightarrow {}^{n+3}C_{n+1} + {}^{2n}C_n = b_n\)
Similarly
\(\Rightarrow {}^{2n}C_{n+1} + {}^{2n}C_n = b_n \Rightarrow {}^{2n+1}C_{n+1} = b_n\) \(\text{H.P.}\)

Question. Find the coefficient of \(x^r\) in the expression of \((x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \dots + (x+2)^{n-1}\).
Answer: Let \((x+3) = a\) & \((x+2) = b\)
Now \(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + b^{n-1}\)
\(= a^{n-1} \left[ 1 + \left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 + \dots + \left(\frac{b}{a}\right)^{n-1} \right]\)
\(= a^{n-1} \left[ \frac{1 - \left(\frac{b}{a}\right)^n}{1 - \frac{b}{a}} \right] = a^{n-1} \frac{a^n - b^n}{a^n} \frac{a}{a - b} = \frac{a^n - b^n}{a - b}\)
\(= \frac{(x+3)^n - (x+2)^n}{(x+3) - (x+2)} = (x+3)^n - (x+2)^n\)
= Coeff. of \(x^r\) in \((x+3)^n\) - coeff. of \(x^r\) in \((x+2)^n\)
\(= {}^nC_r 3^{n-r} - {}^nC_r 2^{n-r} = {}^nC_r (3^{n-r} - 2^{n-r})\)

Question. (a) Find the index n of the binomial \(\left(\frac{x}{5} + \frac{2}{5}\right)^n\) if the 9th term of the expansion has numerically the greatest coefficient (\(n \in N\)).
(b) For which positive values of x is the fourth term in the expansion of \((5 + 3x)^{10}\) is the greatest.
Answer: (a) If we finding coeff. of \(9^{th}\) term i.e. x = 1
\(\Rightarrow \frac{T_9}{T_8} > 1\) & \(\frac{T_9}{T_{10}} > 1\)
\(\Rightarrow \frac{n+1}{| \frac{1 \cdot 5}{5 \cdot 2} | + 1} - 1 < 8 < \frac{n+1}{| \frac{1 \cdot 5}{5 \cdot 2} | + 1}\)
\(\Rightarrow \frac{2(n+1)}{3} - 1 < 8 < \frac{2(n+1)}{3}\)
\(\Rightarrow \frac{2n - 1}{3} < 8 \text{ or } 8 < \frac{2n + 2}{3}\)
\(\Rightarrow 2n - 1 < 24 \text{ or } 24 < 2n + 2\)
\(\Rightarrow n < 12.5 \text{ or } 11 < n\)
\(\Rightarrow 11 < n < 12.5 \Rightarrow n = 12\)
(b) \(\frac{T_4}{T_3} > 1\) & \(\frac{T_4}{T_5} > 1\)
\(\Rightarrow \frac{11}{| \frac{5}{3x} | + 1} - 1 < 3 < \frac{11}{| \frac{5}{3x} | + 1}\)
\(\Rightarrow \frac{11|3x|}{5 + 3|x|} - 1 < 3 < \frac{11|3x|}{5 + 3|x|}\)
\(\Rightarrow \frac{33|x| - 5 - 3|x|}{5 + 3|x|} < 3 \text{ and } 3 < \frac{33|x|}{5 + 3|x|}\)
\(\Rightarrow 30|x| - 5 < 15 + 9|x| \text{ and } 15 + 9|x| < 33|x|\)
\(\Rightarrow 21|x| < 20 \text{ and } 15 < 24|x|\)
\(\Rightarrow |x| < \frac{20}{21} \text{ and } \frac{15}{24} < |x|\)
\(\Rightarrow x \in \left(-\frac{20}{21}, \frac{20}{21}\right) \text{ and } x \in \left(-\infty, -\frac{5}{8}\right) \cup \left(\frac{5}{8}, \infty\right)\)
\(\Rightarrow x \in \left(-\frac{20}{21}, -\frac{5}{8}\right) \cup \left(\frac{5}{8}, \frac{20}{21}\right)\)
For Positive x \(\in \left(\frac{5}{8}, \frac{20}{21}\right)\)

Question. Prove that \(\frac{(72)!}{(36!)^2} - 1\) is divisible by 73.
Answer: \({}^{72}C_{36} - 1 = {}^{73}C_{36} - {}^{72}C_{35} - 1\)
\(\{ \Rightarrow {}^{72}C_{36} + {}^{72}C_{35} = {}^{73}C_{36} \Rightarrow {}^nC_r = {}^{n+1}C_r - {}^nC_{r-1} \}\)
\(= {}^{73}C_{36} - ({}^{73}C_{35} - {}^{72}C_{34}) - 1\)
\(= {}^{73}C_{36} - {}^{73}C_{35} + ({}^{73}C_{34} - {}^{72}C_{33}) - 1\)
\(= {}^{73}C_{36} - {}^{73}C_{35} + {}^{73}C_{34} - {}^{73}C_{33} + \dots + {}^{73}C_2 - {}^{73}C_1 + {}^{73}C_0 - 1\)
\(= 73 \left[ \frac{72!}{36!37!} - \frac{72!}{35!38!} + \frac{72!}{34!39!} - \dots + \frac{72!}{2!7!} - \frac{72!}{1!72!} \right]\)
= Which is divisible by 73.

Question. (a) Find the number of divisors of the number \(N = {}^{2000}C_1 + 2 \cdot {}^{2000}C_2 + 3 \cdot {}^{2000}C_3 + \dots + 2000 \cdot {}^{2000}C_{2000}\)
(b) Find the sum of the roots (real or complex) of the equation \(x^{2001} + \left(\frac{1}{2} - x\right)^{2001} = 0\).
Answer: (a) \(N = 0 \cdot {}^{2000}C_0 + 1 \cdot {}^{2000}C_1 + \dots + 2000 \cdot {}^{2000}C_{2000} \dots(1)\)
\(N = 2000 \cdot {}^{2000}C_0 + 1999 \cdot {}^{2000}C_1 + \dots + 0 \cdot {}^{2000}C_{2000} \dots(2)\)
adding (1) & (2)
\(\Rightarrow 2N = 2000 \cdot [{}^{2000}C_0 + {}^{2000}C_1 + \dots + {}^{2000}C_{2000}]\)
\(\Rightarrow 2N = 2000 \cdot 2^{2000} = 2^4 \cdot 5^3 \cdot 2^{2000} = 2^{2004} \cdot 5^3\)
\(\Rightarrow N = 2^{2003} \cdot 5^3\)
divisors are \(1, 2^1, 2^2, 2^3, \dots, 2^{2003}, 5, 5^2, 5^3\)
\(5 \cdot 2, 5 \cdot 2^2, 5 \cdot 2^3, \dots, 5 \cdot 2^{2003}\) &
\(5^2 \cdot 2, 5^2 \cdot 2^2, 5^2 \cdot 2^3, \dots, 5^2 \cdot 2^{2003}\) &
\(5^3 \cdot 2, 5^3 \cdot 2^2, 5^3 \cdot 2^3, \dots, 5^3 \cdot 2^{2003}\)
\(= 1 + 2003 + 3 + 2003 + 2003 + 2003 = 8016\)
(b) \(x^{2001} + \left(\frac{1}{2} - x\right)^{2001} = 0\)
\(\Rightarrow x^{2001} + {}^{2001}C_0 \left(\frac{1}{2}\right)^{2001} (-x)^0 + {}^{2001}C_1 \left(\frac{1}{2}\right)^{2000} (-x)^1 + \dots + {}^{2001}C_{2000} \left(\frac{1}{2}\right)^1 (-x)^{2000} + {}^{2001}C_{2001} \left(\frac{1}{2}\right)^0 (-x)^{2001} = 0\)
\(\Rightarrow {}^{2001}C_0 \left(\frac{1}{2}\right)^{2001} - {}^{2001}C_1 \left(\frac{1}{2}\right)^{2000} x + \dots - {}^{2001}C_{1999} \left(\frac{1}{2}\right)^2 x^{1999} + {}^{2001}C_{2000} \left(\frac{1}{2}\right)^1 x^{2000} = 0\)
\(\Rightarrow A_1 x^{2000} + A_2 x^{1999} + A_3 x^{1998} + \dots + A_{2000} x + A_{2001} = 0\)
\(A_1 = {}^{2001}C_{2000} \left(\frac{1}{2}\right)\) & \(A_2 = -{}^{2001}C_{1999} \left(\frac{1}{2}\right)^2\)
Sum of roots \(= -\frac{A_2}{A_1} = \frac{\left(\frac{1}{2}\right)^2 {}^{2001}C_{1999}}{\left(\frac{1}{2}\right) {}^{2001}C_{2000}} = \frac{1}{2} \frac{{}^{2001}C_{1999}}{{}^{2001}C_{2000}}\)
\(= \frac{1}{2} \frac{2000}{2} = 500\)

Question. (a) Show that the integral part in each of the following is odd. \(n \in N\).
(i) \((5 + 2\sqrt{6})^n\)
(ii) \((8 + 3\sqrt{7})^n\)
(iii) \((6 + \sqrt{35})^n\)
(b) Show that the integral part in each of the following is even. \(n \in N\).
(i) \((3\sqrt{3} + 5)^{2n+1}\)
(ii) \((5\sqrt{5} + 11)^{2n+1}\)
Answer: (a)(i) Let \((5 + 2\sqrt{6})^n = \text{I} + \text{f}\) where \(0 < \text{f} < 1\)
& \((5 - 2\sqrt{6})^n = \text{f}'\) where \(0 < \text{f}' < 1\) {\(\because 5 > 2\sqrt{6}\)}
\(\Rightarrow \text{I} + \text{f} + \text{f}' = \text{Even integer} \Rightarrow \text{f} + \text{f}' = 1\)
\(\because 0 < \text{f} + \text{f}' < 2\)
\(\Rightarrow \text{I} = \text{even integer} - (\text{f} + \text{f}')\)
\(= \text{even integer} - 1 = \text{odd integer}\)
(ii) Let \((8 + 3\sqrt{7})^n = \text{I} + \text{f}\) where \(0 < \text{f} < 1\)
& \((8 - 3\sqrt{7})^n = \text{f}'\) where \(0 < \text{f}' < 1\) {\(\because 8 > 3\sqrt{7}\)}
\(\Rightarrow \text{I} + \text{f} + \text{f}' = \text{Even integer} \Rightarrow \text{f} + \text{f}' = 1\)
\(\because 0 < \text{f} + \text{f}' < 2\)
\(\Rightarrow \text{I} = \text{even integer} - (\text{f} + \text{f}')\)
\(= \text{even integer} - 1 = \text{odd integer}\)
(iii) Let \((6 + \sqrt{35})^n = \text{I} + \text{f}\) where \(0 < \text{f} < 1\)
& \((6 - \sqrt{35})^n = \text{f}'\) where \(0 < \text{f}' < 1\) {\(\because 6 > \sqrt{35}\)}
\(\Rightarrow \text{I} + \text{f} + \text{f}' = \text{Even integer} \Rightarrow \text{f} + \text{f}' = 1\)
\(\because 0 < \text{f} + \text{f}' < 2\)
\(\Rightarrow \text{I} = \text{even integer} - (\text{f} + \text{f}')\)
\(= \text{even integer} - 1 = \text{odd integer}\)
(b)(i) Let \((3\sqrt{3} + 5)^{2n+1} = \text{I} + \text{f}\) ; \(0 < \text{f} < 1\)
& \((3\sqrt{3} - 5)^{2n+1} = \text{f}'\) ; \(0 < \text{f}' < 1\) {\(\because 3\sqrt{3} > 5\)}
\(\Rightarrow \text{I} + \text{f} - \text{f}' = \text{even integer}\)
\(\Rightarrow 0 < \text{f} < 1\) & \(0 < \text{f}' < 1\)
\(\therefore -1 < -\text{f}' < 0 \Rightarrow -1 < \text{f} - \text{f}' < 1\)
\(\Rightarrow \text{f} - \text{f}' = 0\) \(\because \{\text{f} - \text{f}'\} < 1\)
\(\therefore \text{f} - \text{f}'\) should be integer
\(\Rightarrow \text{I} + 0 = \text{even integer} \Rightarrow \text{I} = \text{even integer}\)
(ii) Let \((5\sqrt{5} + 11)^{2n+1} = \text{I} + \text{f}\) ; \(0 < \text{f} < 1\)
& \((5\sqrt{5} - 11)^{2n+1} = \text{f}'\) ; \(0 < \text{f}' < 1\) {\(\because 5\sqrt{5} > 11\)}
\(\Rightarrow \text{I} + \text{f} - \text{f}' = \text{even integer}\)
\(\Rightarrow 0 < \text{f} < 1\) & \(0 < \text{f}' < 1\)
\(\therefore -1 < -\text{f}' < 0 \Rightarrow -1 < \text{f} - \text{f}' < 1\)
\(\Rightarrow \text{f} - \text{f}' = 0\) \(\because \{\text{f} - \text{f}'\} < 1\)
\(\therefore \text{f} - \text{f}'\) should be integer
\(\Rightarrow \text{I} + 0 = \text{even integer} \Rightarrow \text{I} = \text{even integer}\)

Question. If \((7 + 4\sqrt{3})^n = p + \beta\) where n and p are positive integers and \(\beta\) is a proper fraction show that \((1 - \beta) (p + \beta) = 1\).
Answer: Let \((7 - 4\sqrt{3})^n = b'\) \(\therefore 0 < b' < 1\) \(\because 7 > 4\sqrt{3}\)
\(p + \beta + b' = \text{even integer}\)
\(\beta + b' = \text{integer} \{ \because 0 < \beta < 1, 0 < b' < 1 \}\)
\(\therefore 0 < \beta + b' < 2 \Rightarrow \beta + b' = 1 \Rightarrow b' = (1 - \beta)\)
Now \((1 - \beta) (p + \beta) = b' (p + \beta)\)
\(= (7 - 4\sqrt{3})^n (7 + 4\sqrt{3})^n = (49 - 48)^n = 1\)

Question. If \((6\sqrt{6} + 14)^{2n+1} = N\) and F be the fractional part of N, prove that \(NF = 20^{2n+1} (n \in N)\).
Answer: Let \((6\sqrt{6} + 14)^{2n+1} = I + F\), \(0 < F < 1\)
& \((6\sqrt{6} - 14)^{2n+1} = F'\), \(0 < F' < 1\) {\(\because 6\sqrt{6} > 14\)}
\(\Rightarrow I + F - F' = \text{even integer}\)
\(\Rightarrow N - F' = \text{even integer} \therefore NF = NF'\)
\(= (6\sqrt{6} + 14)^{2n+1} \cdot (6\sqrt{6} - 14)^{2n+1}\)
\(= ((6\sqrt{6})^2 - 14^2)^{2n+1} = (216 - 196)^{2n+1} = 20^{2n+1}\)

Question. Prove that the integer next above \((\sqrt{3} + 1)^{2n}\) contains \(2^{n+1}\) as factor (\(n \in N\)).
Answer: Integer next above \((\sqrt{3} + 1)^{2n}\) contains \(2^{n+1}\) as factor n \(\in\) w i.e. \([(\sqrt{3} + 1)^{2n}] + 1\) contains \(2^{n+1}\) as factor.
Let \((\sqrt{3} + 1)^{2n} = I + f\) \(0 < f < 1\)
& \((\sqrt{3} - 1)^{2n} = f'\) \(0 < f' < 1\)
\(\Rightarrow I + f + f' = I + 1 = \text{even integer}\)
\(= (\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n}\)
\(= \{(\sqrt{3} + 1)^2\}^n + \{(\sqrt{3} - 1)^2\}^n\)
\(= \{4 + 2\sqrt{3}\}^n + \{4 - 2\sqrt{3}\}^n\)
\(= 2^n [(2 + \sqrt{3})^n + (2 - \sqrt{3})^n]\)
\(= 2^n \cdot 2[{}^nC_0 2^m + {}^nC_2 2^{m-2} (\sqrt{3})^2 + \dots]\)
\(= 2^{n+1} [\text{any positive integer}] \)
which is divisible by \(2^{n+1}\)

Question. Let I denotes the integral part and F the proper fractional part of \((3 + \sqrt{5})^n\) where \(n \in N\) and if \(\rho\) denotes the rational part and \(\sigma\) the irrational part of the same, show that \(\rho = \frac{1}{2}(I + 1)\) and \(\sigma = \frac{1}{2}(I + 2F - 1)\).
Answer: Given \((3 + \sqrt{5})^n = I + F \therefore 0 < F < 1, n \in N\)
Let \((3 - \sqrt{5})^n = F'\), \(0 < F' < 1\) (\(\because 3 > \sqrt{5}\))
\(\Rightarrow I + F + F' = \text{even integer} \Rightarrow F + F' = 1\)
Now \(I + F = \rho + \sigma\) .... (i) & \(F' = \rho - \sigma\) ....(ii)
\(I + (F + F') = 2\rho \Rightarrow \rho = \left(\frac{I + 1}{2}\right)\)
Put in (ii) \(\sigma = \frac{1}{2}(I + 2F - 1)\) H.P.

Question. Prove that \(\frac{^{2n}C_n}{n + 1}\) is an integer, \(\forall n \in N\).
Answer: \(\frac{^{2n}C_n}{n + 1}\) is an integer \(\forall n \in N\)
\(= \frac{^{2n}C_n}{(n + 1)} [(2n + 2) - (2n + 1)]\)
\(= {}^{2n}C_n \left[ \frac{2(n + 1)}{(n + 1)} - \frac{(2n + 1)}{(n + 1)} \right]\)
\(= 2 \cdot {}^{2n}C_n - \frac{(2n + 1)}{(n + 1)} {}^{2n}C_n \quad \{ {}^{n+1}C_{r+1} = \frac{(n + 1)}{(r + 1)} {}^nC_r \}\)
\(= 2 \cdot {}^{2n}C_n - {}^{2n+1}C_{n+1}\)
\({}^{2n}C_n\) & \({}^{2n+1}C_{n+1}\) are integer
& subtraction of two integer is also integer.
Aliter :
\(\frac{^{2n}C_n}{n + 1} (n + 1 - n) = {}^{2n}C_n - \frac{n}{n + 1} {}^{2n}C_n\)
\(= {}^{2n}C_n - \frac{n}{n+1} \frac{(2n)!}{n!n!} = {}^{2n}C_n - \frac{2n!}{(n+1)!(n-1)!}\)
\(= {}^{2n}C_n - {}^{2n}C_{n+1} \text{ or } {}^{2n}C_n - {}^{2n}C_{n-1}\)

Advanced Subjective Questions

Question. If \( C_0, C_1, C_2, \dots, C_n \) are the combinatorial coefficients in the expansion of \( (1 + x)^n, n \in N \), then prove the following: \( C_0^2 + C_1^2 + C_2^2 + \dots + C_n^2 = \frac{(2n)!}{n! n!} \) (This result is to be remembered)
Answer: \( (1 + x)^n = C_0 + C_1x^1 + C_2x^2 + \dots + C_rx^r \dots + C_nx^n \dots(i) \)
\( (x + 1)^n = C_0x^n + C_1x^{n-1} + C_2x^{n-2} + \dots + C_nx^0 \dots(ii) \)
Multiply (i) & (ii) and compare the coeff. of \( x^n \) on both sides \( C_0^2 + C_1^2 + C_2^2 + \dots + C_n^2 = \text{coeff. of } x^n \text{ in } (1 + x)^{2n} = {}^{2n}C_n = \frac{(2n)!}{(n!)(n!)} \)

Question. \( C_0 C_1 + C_1 C_2 + C_2 C_3 + \dots + C_{n - 1} C_n = \frac{(2n)!}{(n + 1)! (n - 1)!} \)
Answer: Multiply (i) & (ii) and compare the coeff. of \( x^{n+1} \) or \( x^{n-1} \) both side \( C_0C_1 + C_1C_2 + \dots + C_{n-1} C_n = \text{coeff. of } x^{n+1} \text{ or } x^{n-1} \text{ in } (1+ x)^{2n} = {}^{2n}C_{n+1} = {}^{2n}C_{n-1} = \frac{(2n)!}{(n + 1)!(n - 1)!} \)

Question. \( C_1 + 2C_2 + 3C_3 + \dots + n \cdot C_n = n \cdot 2^{n - 1} \)
Answer: Let \( S = 0 \cdot C_0 + 1 \cdot C_1 + 2 \cdot C_2 + 3 \cdot C_3 + \dots + nC_n \)
\( S = n \cdot C_0 + (n - 1) C_1 + (n - 2) C_2 + \dots 1 \cdot C_{n-1} + 0C_n \)
\( 2S = n (C_0 + C_1 + C_2 + \dots + C_n) \)
\( \Rightarrow 2S = n \cdot 2^n \Rightarrow S = n \cdot 2^{n-1} \)
Aliter :
\( \sum T_p = \sum_{p=1}^n p \cdot {}^nC_p = \sum p \cdot \frac{n}{p} {}^{n-1}C_{p-1} = n \sum_{p=1}^n {}^{n-1}C_{p-1} \)
\( S = n [^{n-1}C_0 + {}^{n-1}C_1 + \dots + {}^{n-1}C_{n-1}] = n \cdot 2^{n-1} \)
Aliter :
\( (1 + x)^n = C_0 + C_1x + C_2x^2 + C_3x^3 + \dots + C_nx^n \) diff. w.r.t to \( x \)
\( n(1 + x)^{n-1} = C_1 + 2C_2x + 3C_3 x^2 + \dots + n C_n x^{n-1} \)
Put \( x = 1 \)
\( C_1 + 2C_2 + 3 \cdot C_3 + \dots + nC_n = n (1 + 1)^{n-1} = n \cdot 2^{n-1} \)

Question. \( C_0 + 2C_1 + 3C_2 + \dots + (n + 1)C_n = (n + 2)2^{n - 1} \)
Answer: \( S = C_0 + 2C_1 + 3C_2 + \dots + (n + 1) C_n \)
\( S = (n + 1) C_0 + (n) C_1 + (n - 1) C_2 + \dots + 1C_n \)
\( \Rightarrow 2S = (n + 2) [C_0 + C_1 + C_2 + \dots + C_n] = (n + 2)2^n \)
\( \Rightarrow S = (n + 2) \cdot 2^{n-1} \)
Aliter : \( \sum T_{P+1} = \sum_{P=0}^n (P + 1)C_P = \sum PC_P + \sum C_P = n \cdot 2^{n-1} + 2^n = 2^{n-1} [n + 2] \)

Question. \( C_0 + 3C_1 + 5C_2 + \dots + (2n + 1)C_n = (n + 1) 2^n \)
Answer: \( S = C_0 + 3C_1 + 5C_2 + \dots + (2n + 1) C_n \)
\( S = (2n + 1) C_0 + (2n - 1) C_1 + \dots + C_n \)
\( 2S = (2n + 2) [C_0 + C_1 + \dots C_n] = 2 (n + 1) 2^n \)
\( \Rightarrow S = (n + 1) 2^n \)
Aliter : \( T_{p+1} = \sum_{P=0}^n (2P + 1)C_P = 2\sum P \cdot C_P + \sum C_P = 2 \cdot n \cdot 2^{n-1} + 2^n = 2^n (n + 1) \)

Question. \( (C_0 + C_1) (C_1 + C_2) (C_2 + C_3) \dots (C_{n - 1} + C_n) = \frac{C_0 \cdot C_1 \cdot C_2 \dots C_{n-1} (n + 1)^n}{n!} \)
Answer: L.H.S. \( = {}^{n+1}C_1 \cdot {}^{n+1}C_2 \cdot {}^{n+1}C_3 \dots {}^{n+1}C_n = \frac{(n + 1)}{1} C_0 \cdot \frac{(n + 1)}{2} C_1 \cdot \frac{(n + 1)}{3} C_2 \dots \frac{(n + 1)}{n} C_{n-1} = \frac{C_0 \cdot C_1 \cdot C_2 \dots C_{n-1}}{n!} (n + 1)^n = \text{R.H.S.} \)

Question. If \( P_n \) denotes the product of all the coefficients in the expansion of \( (1 + x)^n, n \in N \), show that, \( \frac{P_{n+1}}{P_n} = \frac{(n+1)^n}{n!} \)
Answer: Given \( P_n = C_0 \cdot C_1 \cdot C_2 \dots C_{n-1} C_n \)
\( \therefore \frac{P_{n+1}}{P_n} = \frac{{}^{n+1}C_0 \cdot {}^{n+1}C_1 \cdot {}^{n+1}C_2 \dots {}^{n+1}C_n \cdot {}^{n+1}C_{n+1}}{{}^nC_0 \cdot {}^nC_1 \cdot {}^nC_2 \dots {}^nC_{n-1} \cdot {}^nC_n} \)
\( \{^{n+1}C_{n+1} = 1 = {}^nC_n; {}^{n+1}C_0 = {}^nC_0 = 1\} \)
\( \Rightarrow \frac{P_{n+1}}{P_n} = \frac{{}^{n+1}C_1 \cdot {}^{n+1}C_2 \dots {}^{n+1}C_n}{C_0 \cdot C_1 \dots C_{n-1}} = \frac{\frac{(n+1)}{1} C_0 \cdot \frac{(n+1)}{2} C_1 \dots \frac{(n+1)}{n} C_{n-1}}{C_0 \cdot C_1 \cdot C_2 \dots C_{n-1}} \)
\( \Rightarrow \frac{P_{n+1}}{P_n} = \frac{(n + 1)^n}{n!} \)

Question. \( \frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + \dots + \frac{n \cdot C_n}{C_{n-1}} = \frac{n(n + 1)}{2} \)
Answer: L.H.S. \( = \left( \frac{n - 0}{1} \right) + 2 \left( \frac{n - 1}{2} \right) + 3 \left( \frac{n - 2}{3} \right) + \dots + n \left( \frac{n - (n - 1)}{n} \right) = n + (n - 1) + (n - 2) + \dots + [n - (n - 1)] = n^2 - [1 + 2 + 3 + \dots + (n - 1)] = n^2 - \frac{n(n - 1)}{2} = \frac{2n^2 - n^2 + n}{2} = \frac{n(n + 1)}{2} = \text{R.H.S} \)

Question. \( C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \dots + \frac{C_n}{n + 1} = \frac{2^{n+1} - 1}{n + 1} \)
Answer: L.H.S. \( = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \dots + \frac{C_n}{(n + 1)} \)
\( \sum T_{P+1} = \sum_{p=0}^n \frac{C_p}{P + 1} = \sum_{p=0}^n \frac{{}^nC_p}{P+1} \times \frac{(n+1)}{(n+1)} = \frac{1}{(n+1)} \sum_{p=0}^n \frac{(n+1)}{p+1} \cdot {}^nC_p = \frac{1}{(n+1)} \sum_{0}^n {}^{n+1}C_{p+1} \)
\( = \frac{1}{(n+1)} [{}^{n+1}C_1 + {}^{n+1}C_2 + \dots + {}^{n+1}C_{n+1}] \)
\( = \frac{1}{(n+1)} [2^{n+1} - {}^{n+1}C_0] = \frac{1}{(n+1)} (2^{n+1} - 1) = \text{R.H.S.} \)
Aliter : \( (1 + x)^n = C_0 + C_1x + C_2x^2 + \dots + C_nx^n \)
Integration w.r. to \( x \)
\( \frac{(1 + x)^{n+1}}{n + 1} + k = C_0x + C_1 \frac{x^2}{2} + C_2 \frac{x^3}{3} + \dots + C_n \frac{x^{n+1}}{n + 1} \)
where k is integral coeff. If \( x = 0 \Rightarrow k = -\frac{1}{(n + 1)} \)
\( \frac{(1 + x)^{n+1}}{(n + 1)} - \frac{1}{(n + 1)} = C_0x + \frac{C_1}{2} x^2 + \frac{C_2}{3} x^3 + \dots + \frac{C_n}{n + 1} x^{n+1} \dots(iii) \)
Put \( x = 1 \), \( \frac{(2^{n+1} - 1)}{(n + 1)} = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \frac{C_3}{4} + \dots + \frac{C_n}{n + 1} \)

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

Question. \( C_0C_r + C_1C_{r + 1} + C_2C_{r + 2} + \dots + C_{n - r}C_n = \frac{2n!}{(n - r)!(n + r)!} \)
Answer: Multiply \( (1+x)^n \) and \( (x+1)^n \) and compare of coeff. of \( x^{n-r} \) or \( x^{n+r} \)
\( C_0C_r + C_1C_{r+1} + C_2C_{r+2} + \dots + C_{n-r}C_n = x^{n+r} \text{ in } (1+x)^{2n} = {}^{2n}C_{n+r} = {}^{2n}C_{n-r} = \frac{(2n)!}{(n - r)!(n + r)!} \)

JEE Problems

Question. Prove that : \( 2^k \binom{n}{0}\binom{n}{k} - 2^{k-1}\binom{n}{1}\binom{n-1}{k-1} + 2^{k-2}\binom{n}{2}\binom{n-2}{k-2} - \dots + (-1)^k \binom{n}{k}\binom{n-k}{0} = \binom{n}{k} \)
Answer: The general term is \( T_{r+1} = (-1)^r \binom{n}{r} \binom{n-r}{k-r} 2^{k-r} \).
Expanding the binomial coefficients: \( \frac{n!}{r!(n-r)!} \frac{(n-r)!}{(k-r)!(n-k)!} = \frac{n!}{k!(n-k)!} \frac{k!}{r!(k-r)!} = \binom{n}{k} \binom{k}{r} \).
Summing over \( r \): \( \sum_{r=0}^k \binom{n}{k} \binom{k}{r} (-1)^r 2^{k-r} = \binom{n}{k} \sum_{r=0}^k \binom{k}{r} (-1)^r 2^{k-r} \).
Using binomial theorem \( (2-1)^k = 1^k = 1 \), the sum is \( \binom{n}{k} \cdot 1 = \binom{n}{k} \).

Question. \( {}^{n-1}C_r = (k^2 - 3) \cdot {}^nC_{r+1} \), if \( k \in \)
(a) \( [-\sqrt{3}, \sqrt{3}] \)
(b) \( (-\infty, -2) \)
(c) \( (2, \infty) \)
(d) \( (\sqrt{3}, 2] \)
Answer: (d) \( (\sqrt{3}, 2] \)

Question. The value of \( \binom{30}{0}\binom{30}{10} - \binom{30}{1}\binom{30}{11} + \binom{30}{2}\binom{30}{12} - \dots + \binom{30}{20}\binom{30}{30} \) is, where \( \binom{n}{r} = {}^nC_r \)
(a) \( \binom{30}{10} \)
(b) \( \binom{30}{15} \)
(c) \( \binom{60}{30} \)
(d) \( \binom{31}{10} \)
Answer: (a) \( \binom{30}{10} \)

Question. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is
(a) 55
(b) 66
(c) 77
(d) 88
Answer: (c) 77

Question. For \( r = 0, 1, \dots, 10 \) let \( A_r, B_r, C_r \) denote, respectively, the coefficient of \( x^r \) in the expansions of \( (1 + x)^{10}, (1 + x)^{20} \) and \( (1 + x)^{30} \). Then \( \sum_{r=1}^{10} A_r(B_{10}B_r - C_{10}A_r) \) is equal to
(a) \( B_{10} - C_{10} \)
(b) \( A_{10}(B_{10}^2 - C_{10}A_{10}) \)
(c) 0
(d) \( C_{10} - B_{10} \)
Answer: (d) \( C_{10} - B_{10} \)

Question. Which of the following is correct ?
(a) \( a_{17} = a_{16} + a_{15} \)
(b) \( c_{17} \neq c_{16} + c_{15} \)
(c) \( b_{17} \neq b_{16} + c_{16} \)
(d) \( a_{17} = c_{17} + b_{16} \)
Answer: (a) \( a_{17} = a_{16} + a_{15} \)

Question. The value of \( b_6 \) is
(a) 7
(b) 8
(c) 9
(d) 11
Answer: (b) 8