JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 11)
Let $$m$$ be the smallest positive integer such that the coefficient of $${x^2}$$ in the expansion of $${\left( {1 + x} \right)^2} + {\left( {1 + x} \right)^3} + ........ + {\left( {1 + x} \right)^{49}} + {\left( {1 + mx} \right)^{50}}\,\,$$ is $$\left( {3n + 1} \right)\,{}^{51}{C_3}$$ for some positive integer $$n$$. Then the value of $$n$$ is
Answer
5
Explanation
It is given that the coefficient of $x^2$ in $(1+x)^2+(1+x)^3+$ $\cdots+(1+x)^{49}+(1+m x)^{50}$ is $(3 n+1){ }^{51} C_3$.
Now,
$$ \begin{aligned} & { }^2 C_2+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^3 C_3+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \quad \quad\left(\text { as } n C r+n C r-1=n^{+1} C r\right) \\\\ & \Rightarrow{ }^4 C_3+{ }^4 C_2+\cdots+{ }^{49} C_2+m^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{49} C_3+{ }^{49} C_2+m{ }^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+{ }^{50} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{51} C_3+m^2{ }^{50} C_2=3 n{ }^{51} C_3+{ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{-50} C_2+m^{250} C_2=3 n{ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_2\left(m^2-1\right)=3 n \cdot \frac{51}{3}{ }^{50} C_2\left({ }^n C_r=\frac{n}{r} \quad{ }^{n-1} C_{r-1}\right) \\\\ & \Rightarrow 2\left(m^2-1\right)=51 \end{aligned} $$
Therefore, $m^2=51 n+1$.
$$ \begin{gathered} \Rightarrow n=\frac{m^2-1}{51} \\\\ \Rightarrow m=16 \Rightarrow n=5\left(m, n \in \mathrm{I}^{+}\right) \end{gathered} $$
Now,
$$ \begin{aligned} & { }^2 C_2+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^3 C_3+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \quad \quad\left(\text { as } n C r+n C r-1=n^{+1} C r\right) \\\\ & \Rightarrow{ }^4 C_3+{ }^4 C_2+\cdots+{ }^{49} C_2+m^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{49} C_3+{ }^{49} C_2+m{ }^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+{ }^{50} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{51} C_3+m^2{ }^{50} C_2=3 n{ }^{51} C_3+{ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{-50} C_2+m^{250} C_2=3 n{ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_2\left(m^2-1\right)=3 n \cdot \frac{51}{3}{ }^{50} C_2\left({ }^n C_r=\frac{n}{r} \quad{ }^{n-1} C_{r-1}\right) \\\\ & \Rightarrow 2\left(m^2-1\right)=51 \end{aligned} $$
Therefore, $m^2=51 n+1$.
$$ \begin{gathered} \Rightarrow n=\frac{m^2-1}{51} \\\\ \Rightarrow m=16 \Rightarrow n=5\left(m, n \in \mathrm{I}^{+}\right) \end{gathered} $$
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