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]

Sol. (D)

$\mathrm{A}_{\mathrm{r}}=^{10} \mathrm{C}_{\mathrm{r}}, \quad \mathrm{B}_{\mathrm{r}}=^{20} \mathrm{C}_{\mathrm{r}}, \mathrm{C}_{\mathrm{r}}=^{30} \mathrm{C}_{\mathrm{r}}$

$\sum_{\mathrm{r}=1}^{10}\left(^{20} \mathrm{C}_{10}^{10} \mathrm{C}_{\mathrm{r}}^{20} \mathrm{C}_{\mathrm{r}}-^{30} \mathrm{C}_{10}\left(^{10} \mathrm{C}_{\mathrm{r}}\right)^{2}\right)$

$=^{20} \mathrm{C}_{10}\left(^{10} \mathrm{C}_{1}^{20} \mathrm{C}_{1}+^{10} \mathrm{C}_{2}^{20} \mathrm{C}_{2}+\ldots .+^{10} \mathrm{C}_{10}^{20}\right)$

$=^{20} \mathrm{C}_{10}\left(^{30} \mathrm{C}_{10}^{2}+1\right)-^{30} \mathrm{C}_{10}\left(^{20} \mathrm{C}_{10}-1\right)$

$=^{20} \mathrm{C}_{10}-^{20} \mathrm{C}_{10}=\mathrm{C}_{10}-\mathrm{B}_{10}$


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]

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


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)]

Sol. (C)

Coefficient of $\mathrm{x}^{11}$ in $\left(1+\mathrm{x}^{2}\right)^{4}\left(1+\mathrm{x}^{3}\right)^{7}\left(1+\mathrm{x}^{4}\right)^{12}$

$=^{4} \mathrm{C}_{0} \cdot^{7} \mathrm{C}_{1}^{12} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \cdot^{7} \mathrm{C}_{3}^{12} \mathrm{C}_{0}+^{4} \mathrm{C}_{2} \cdot^{7} \mathrm{C}_{1} \cdot^{12} \mathrm{C}_{1}$

$+^{4} \mathrm{C}_{4} \cdot^{7} \mathrm{C}_{1} \cdot^{12} \mathrm{C}_{0}$

$=462+140+504+7=1113$


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]

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


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]

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. $\Rightarrow 5$


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)]

Sol. 646


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