# Binomial Theorem - JEE Advanced Previous Year Questions with Solutions

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Q. 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}$[JEE 2010, 5] Ans. (D)$\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}$Q. The coefficients of three consecutive terms of$(1+x)^{n+5}$are in the ratio$5: 10: 14 .$Then$\begin{array}{lll}{\text { n }=} & {\text { [JEE (Advanced) } 2013,4 \mathrm{M},-1 \mathrm{M}}\end{array}$[JEE (Advanced) 2013, 4M, –1M] Ans. 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$12r = 5n + 18 ........(2) Solving (1) and (2), we get n = 6 Q. Coefficient of$x^{11}$in the expansion of$\left(1+x^{2}\right)^{4}\left(1+x^{3}\right)^{7}\left(1+x^{4}\right)^{12}$is - [JEE(Advanced)-2014, 3(–1)] Ans. (C) Coefficient of$\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$Q. The coefficient of$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 [JEE 2015, 4M, –0M] Ans. 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 Q. 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 [JEE (Advanced) 2016] Ans. 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.$\Rightarrow 5$Q. 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 [JEE(Advanced)-2018, 3(0)]
Ans. 646

lucky
Sept. 5, 2021, 9:58 p.m.
I clear my doubt ......
Issac Newton
May 2, 2021, 4:40 p.m.
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D
Feb. 1, 2021, 8:30 a.m.
Waste
Arun kumar thulla
Sept. 22, 2020, 3:04 p.m.
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Abhiram
Sept. 19, 2020, 2:19 p.m.
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Vamshi
June 4, 2020, 12:34 p.m.
Sir please send the PDF of binomial theorem chapter and it's problems.🤲🤲
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