Binomial Theorem 1 Question 34
36. Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1+x)^{2}+(1+x)^{3}+\ldots+(1+x)^{49}+(1+m x)^{50}$ is $(3 n+1)$ ${ }^{51} C _3$ for some positive integer $n$. Then, the value of $n$ is
(2016 Adv.)
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Solution:
- Coefficient of $x^{2}$ in the expansion of
$$ {(1+x)^{2}+(1+x)^{3}+\ldots+(1+x)^{49}+(1+m x)^{50} } $$
$\Rightarrow{ }^{2} C _2+{ }^{3} C _2+{ }^{4} C _2+\ldots+{ }^{49} C _2+{ }^{50} C _2 \cdot m^{2}$
$$ \begin{array}{r} \Rightarrow{ }^{2} C _2+{ }^{3} C _2+{ }^{4} C _2+\ldots+{ }^{49} C _2+{ }^{50} C _2 \cdot m^{2} \\ =(3 n+1) \cdot{ }^{51} C _3 \end{array} $$
$$ \begin{array}{r} \Rightarrow \quad{ }^{50} C _3+{ }^{50} C _2 m^{2}=(3 n+1) \cdot{ }^{51} C _3 \\ {\left[\because{ }^{r} C _r+{ }^{r+1} C _r+\ldots+{ }^{n} C _r={ }^{n+1} C _{r+1}\right]} \end{array} $$
$\Rightarrow \quad \frac{50 \times 49 \times 48}{3 \times 2 \times 1}+\frac{50 \times 49}{2} \times m^{2}=(3 n+1) \frac{51 \times 50 \times 49}{3 \times 2 \times 1}$
$\Rightarrow \quad m^{2}=51 n+1$
$\therefore$ Minimum value of $m^{2}$ for which $(51 n+1)$ is integer (perfect square) for $n=5$.
$$ \begin{array}{ll} \therefore & m^{2}=51 \times 5+1 \Rightarrow m^{2}=256 \\ \therefore & m=16 \text { and } n=5 \end{array} $$
Hence, the value of $n$ is 5 .
