Binomial Theorem - JEE Advanced Previous Year Questions with Solutions

eSaral › JEE › JEE Main ›Binomial Theorem - JEE Advanced Previous Year Questions with Solutions

🚀 Checkout eSaral Courses

 Download eSaral app for free study material and video tutorials. [esquestion] For $\mathrm{r}=0,1, \ldots, 10,$ let $\mathrm{A}_{\mathrm{r}}, \mathrm{B}_{\mathrm{r}}$ and $\mathrm{C}_{\mathrm{r}}$ denote, respectively, the coefficient of $\mathrm{x}^{\mathrm{r}}$ in the expansions of $(1+\mathrm{x})^{10},(1+\mathrm{x})^{20}$ and $(1+\mathrm{x})^{30} .$ Then $\sum_{\mathrm{r}=1}^{10} \mathrm{A}_{\mathrm{r}}\left(\mathrm{B}_{10} \mathrm{B}_{\mathrm{r}}-\mathrm{C}_{10} \mathrm{A}_{\mathrm{r}}\right)$ is equal to $-\quad (A) $\mathrm{B}_{10}-\mathrm{C}_{10}$ (B) $\mathrm{A}_{10}\left(\mathrm{B}_{10}^{2}-\mathrm{C}_{10} \mathrm{A}_{10}\right)$ (C) 0 (D) $\mathrm{C}_{10}-\mathrm{B}_{10}$ #tag# [JEE 2010, 5] #sol# (D) $\mathrm{A}_{\mathrm{r}}=^{10} \mathrm{C}_{\mathrm{r}}, \quad \mathrm{B}_{\mathrm{r}}=^{20} \mathrm{C}_{\mathrm{r}}, \mathrm{C}_{\mathrm{r}}=^{30} \mathrm{C}_{\mathrm{r}}$ $\sum_{\mathrm{r}=1}^{10}\left(^{20} \mathrm{C}_{10}^{10} \mathrm{C}_{\mathrm{r}}^{20} \mathrm{C}_{\mathrm{r}}-^{30} \mathrm{C}_{10}\left(^{10} \mathrm{C}_{\mathrm{r}}\right)^{2}\right)$ $=^{20} \mathrm{C}_{10}\left(^{10} \mathrm{C}_{1}^{20} \mathrm{C}_{1}+^{10} \mathrm{C}_{2}^{20} \mathrm{C}_{2}+\ldots .+^{10} \mathrm{C}_{10}^{20}\right)$ $=^{20} \mathrm{C}_{10}\left(^{30} \mathrm{C}_{10}^{2}+1\right)-^{30} \mathrm{C}_{10}\left(^{20} \mathrm{C}_{10}-1\right)$ $=^{20} \mathrm{C}_{10}-^{20} \mathrm{C}_{10}=\mathrm{C}_{10}-\mathrm{B}_{10}$ [/esquestion] [esquestion] The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14 .$ Then $\begin{array}{lll}{\text { n }=} & {\text { [JEE (Advanced) } 2013,4 \mathrm{M},-1 \mathrm{M}}\end{array}$ #tag# [JEE (Advanced) 2013, 4M, –1M] #sol# 6 Let the three consecutive terms be $^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}-1},^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}},^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}+1}$ $\therefore \quad \frac{^{n+5} C_{r-1}}{^{n+5} C_{r}}=\frac{1}{2} \quad \Rightarrow \quad \frac{r}{n-r+6}=\frac{1}{2}$ $\Rightarrow \quad \mathrm{n}=3 \mathrm{r}-6$ ........(1) Also, $\frac{n+5 \mathrm{C}_{\mathrm{r}}}{^{\mathrm{n}+5} \mathrm{C}_{\mathrm{r}+1}}=\frac{5}{7} \Rightarrow \frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}+5}=\frac{5}{7}$. $\Rightarrow$ 12r = 5n + 18 ........(2) Solving (1) and (2), we get n = 6 [/esquestion] [esquestion] Coefficient of $x^{11}$ in the expansion of $\left(1+x^{2}\right)^{4}\left(1+x^{3}\right)^{7}\left(1+x^{4}\right)^{12}$ is - #tag# [JEE(Advanced)-2014, 3(–1)] #sol# (C) Coefficient of $\mathrm{x}^{11}$ in $\left(1+\mathrm{x}^{2}\right)^{4}\left(1+\mathrm{x}^{3}\right)^{7}\left(1+\mathrm{x}^{4}\right)^{12}$ $=^{4} \mathrm{C}_{0} \cdot^{7} \mathrm{C}_{1}^{12} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \cdot^{7} \mathrm{C}_{3}^{12} \mathrm{C}_{0}+^{4} \mathrm{C}_{2} \cdot^{7} \mathrm{C}_{1} \cdot^{12} \mathrm{C}_{1}$ $+^{4} \mathrm{C}_{4} \cdot^{7} \mathrm{C}_{1} \cdot^{12} \mathrm{C}_{0}$ $=462+140+504+7=1113$ [/esquestion] [esquestion] The coefficient of $x^{9}$ in the expansion of $(1+x)\left(1+x^{2}\right)\left(1+x^{3}\right) \ldots\left(1+x^{100}\right)$ is #tag# [JEE 2015, 4M, –0M] #sol# 8 There are 8 product $1^{99} \mathrm{x}^{9}, 1^{98} \mathrm{x}^{8}, 1^{98} \mathrm{x}^{2} \mathrm{x}^{7}, 1^{98} \mathrm{x}^{3} \mathrm{x}^{6}, 1^{98} \mathrm{x}^{4} \mathrm{x}^{5}$ $1^{97} \mathrm{x} \mathrm{x}^{2} \mathrm{x}^{6}, 1^{97} \mathrm{x}^{3} \mathrm{x}^{5}, 1^{97} \mathrm{x}^{2} \mathrm{x}^{3} \mathrm{x}^{4}$ which generate $\mathrm{x}^{9}$ so coeff. is 8 [/esquestion] [esquestion] Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1+x)^{2}+(1+x)^{3}+\ldots \ldots+(1+x)^{49}+(1+m x)^{50}$ is $(3 n+1)^{51} C_{3}$ for some positive integer n. Then the value of $n$ is #tag# [JEE (Advanced) 2016] #sol# 5 Coefficient of $x^{2}$ in the expansion of $(1+x)^{2}+(1+x)^{3}+\ldots \ldots(1+x)^{49}+(1+m x)^{50}$ is $^{2} C_{2}+^{3} C_{2}+\ldots \ldots^{49} C_{2}+^{50} C_{2} m^{2}=(3 n+1)^{51} C_{3}$ $^{3} C_{3}+^{3} C_{2}+\ldots \ldots^{49} C_{2}+^{50} C_{2} m^{2}=(3 n+1)^{51} C_{3}$ $\quad^{50} C_{3}+^{50} C_{2} m^{2}=(3 n+1)^{51} C_{3}$ $\frac{50.49 .48}{6}+\frac{50.49}{2} \mathrm{m}^{2}=(3 \mathrm{n}+1) \frac{51.50 .49}{6}$ $\mathrm{m}^{2}=51 \mathrm{n}+1$ must be a perfect square $\Rightarrow \mathrm{n}=5$ and $\mathrm{m}=16$ Ans. $\Rightarrow 5$ [/esquestion] [esquestion] Let $\mathrm{X}=\left(^{10} \mathrm{C}_{1}\right)^{2}+2\left(^{10} \mathrm{C}_{2}\right)^{2}+3\left(^{10} \mathrm{C}_{3}\right)^{2}+\ldots+10\left(^{10} \mathrm{C}_{10}\right)^{2},$ where $^{10} \mathrm{C}_{\mathrm{r}^{\prime}}, \mathrm{r} \in\{1,2, \ldots, 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} \mathrm{X}$ is #tag# [JEE(Advanced)-2018, 3(0)] #sol# 646 [/esquestion]

The Binomial Theorem - JEE Advanced PYQs (Maths) set can be followed by original papers from the JEE Advanced PYQ archive.