First slide

Binomial theorem for positive integral Index

Question

Let $n$ be the smallest positive integer such that the coefficients $x^{2}$ in the expansion $(1 + x)^{2} + (1 + x)^{3} + \dots + (1 + x)^{49} + (1 + m x)^{50}$ is $(3 n + 1)^{51} C_{3}$ for some positive integer $n$ . Then the value of $n$ , is

Moderate

A

16
B

5
C

21
D

11

Solution

We find that
$\begin{array}{l} (1 + x)^{2} + (1 + x)^{3} + \dots (1 + 1)^{40} + (1 + m x)^{50} \\ = (1 + x)^{2} \{\frac{(1 + x)^{48} - 1}{(1 + x) - 1}] + (1 + m x)^{30} \\ = \frac{1}{x} \{(1 + x)^{50} - (1 + x)^{2}\} + (1 + m x)^{30} \end{array}$
$∴$ Coeffi. of $x^{2}$ in $(1 + x)^{2} + (1 + x)^{3} + \dots (1 + x)^{49} + (1 + m x)^{50}$
= Coeffi. of $x^{2}$ in $[\frac{1}{x} \{(1 + x)^{30} - (1 + x)^{2}\} + (1 + m x)^{50}]$
= Coeffi. of $x^{3}$ in $\{(1 + x)^{50} - (1 + x)^{2}\} +$ coefficient of $x^{2}$ in $(1 + m x)^{50}$
$=^{50} C_{3} +^{51} C_{2} m^{2}$
It is given that this coefficient is $(3 n + 1)^{51} C_{3}$
$\begin{array}{l} ∴^{50} C_{3} +^{50} C_{2} m^{2} = (3 n + 1)^{51} C_{3} \\ \Rightarrow \frac{50 \times 49 \times 48}{3 \times 2 \times 1} + \frac{50 \times 49}{2 \times 1} m^{2} = (3 n + 1) \frac{51 \times 50 \times 49}{3 \times 2 \times 1} \\ \Rightarrow 16 + m^{2} = 17 (3 n + 1) \\ \Rightarrow m^{2} = 51 n + 1 \end{array}$
The least value of n for which $51 n + 1$ is a perfect square is 5 and for this value of $n$ the value of $m$ is $16$ . Hence,
$m = 16$ and $n = 5 .$

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