Question

Let m be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x{)}^2+(1+x{)}^3+\cdots +(1+x{)}^{49}+(1+mx{)}^{50}is(3n+1{)}^{51}{C}_3$ for some positive integer n. Then, the value of n is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The correct option is B 5

Coefficient of $x^2$ in the expansion of
$\left\{\right(1+x{)}^2+(1+x{)}^3+\cdots +(1+x{)}^{49}+(1+mx{)}^{50}\}\Rightarrow {}^2{C}_2{+}^3{C}_2{+}^4{C}_2+\cdots {+}^{49}{C}_2{+}^{50}{C}_2.m^2=(3n+1){.}^{51}{C}_3\Rightarrow {}^{50}{C}_3{+}^{50}{C}_2{m}^2=(3n+1){.}^{51}{C}_3[{\therefore }^r{C}_r{+}^{r+1}{C}_r+\cdots {+}^n{C}_r{=}^{n+1}{C}_{r+1}]$
$\Rightarrow \frac{50\times 49\times 48}{3\times 2\times 1}+\frac{50\times 49}{2}\times m^2=(3n+1)\frac{51\times 50\times 49}{3\times 2\times 1}\Rightarrow m^2=51n+1$
$\therefore$ Minimum value of $m^2$ for which (51n + 1) is integer (perfect square) for n = 5.
$\therefore m^2=51\times 5+1\Rightarrow m^2=256$
$\therefore$ m = 16 and n = 5
Hence, the value of n is 5.