Question

The coefficeint of $x^r$ in the expansion of ${\left(1+3x+6x^2+10x^3+......\right)}^{2}is:$

• A. $\frac{\left(r+1\right)\left(r+2\right)(r+3)}{5!}$
• B. $\frac{\left(r+2\right)\left(r+3\right)(r+4)}{5!}$
• C. $\frac{\left(r+1\right)\left(r+2\right)\left(r+3\right)(r+4)(r+5)}{5!}$
• D. none of these

Answer 1

The correct option is A $\frac{\left(r+1\right)\left(r+2\right)(r+3)}{5!}$

$(1+3x+6x^2+10x^3........+\infty )$

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3$

or

$(1+y{)}^n=1+ny+\frac{n(n-1)y^2}{2!}+\frac{n(n-1)(n-2)}{3!}{y}^3$

$ny=3x\frac{n(n-1)y^2}{2!}=6x^2$

$y=\frac{3x}{n}\frac{n(n-1)}{2}\times \frac{9x^2}{n^2}=6x^2$

$3n-3=4n$

$n=-3$

$\left(\right(1+x{)}^{-3}{)}^2=(1+x{)}^{-6}$

$=(1+x{)}^{-6}=1+2x+\frac{2.3}{2b}{x}^2+\frac{\mathrm{2.3.}y}{3!}{x}^3$

${=}^6{C}_r\left(x{)}^r\right(1{)}^{6-r}$

$=\frac{(r+1)(r+2)(r+3)}{5!}$