Question

The coefficient of x${}^{24}$ in the expansion of
(1 +3x + 6x${}^2$ + 10x${}^3$+ -----------+$\infty$)${}^{2/3}$ =

• A. 300
• B. 250
• C. 25
• D. 205

Answer 1

Answer: (C)

25

${(1+3x+6x^2+10x^3+\dots \infty )}^{\frac{2}{3}}$

$\text{We know that,}$

${(1+\alpha )}^n=1+n\alpha +\frac{n(n-1){\alpha }^2}{2!}+\dots$

$\text{comparing with given eqn,}$

$\frac{n(n-1)}{2}\cdot {\alpha }^2=6x^2$

$=n(n-1)\frac{9x^2}{n^2}=6x^2$

$\Rightarrow n=-3$

$n\alpha =3x$

$\Rightarrow \alpha =\frac{3x}{n}$

$\alpha =-x$

$\therefore \text{Expansion is}(1-x{)}^{-\text{/}3\times \frac{2}{\text{/}3}}$

$=(1-x{)}^{-2}$

$=1+2x+\frac{2\times 3}{2!}{x}^2+\frac{2\times 3\times 4}{3!}{x}^3+\dots$

$\therefore$ coeff. of $x^{24}$ is 2

$=\frac{2\times 3\times 4\times \dots 25}{24!}$

$=25$