Binomial Theorem Ques 34
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.)
Show Answer
Answer:
Correct Answer: 34.$(5)$
Solution:
Formula:
Important results on binomial coefficients :
- 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} $
$\Rightarrow { }^{50} C_{3}+{ }^{50} C_{2} m^{2}=(3 n+1) \cdot{ }^{51} C_{3} $
${\left[\because{ }^{r} C_{r}+{ }^{r+1} C_{r}+\ldots+{ }^{n} C_{r}={ }^{n+1} C_{r+1}\right]}$
$\Rightarrow \quad \frac{50 \times 49 \times 48}{3 \times 2 \times 1}+\frac{50 \times 49}{2} \times m^{2}=(3 n+1) \frac{51 \times 50 \times 49}{3 \times 2 \times 1}$
$\Rightarrow \quad m^{2}=51 n+1$
$\therefore$ 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 $5$ .
BETA
