Question

The sum of the series $1+3x+6x^2+10x^3+...\infty$ is

• A. $\frac{1}{(1-x{)}^2}$
• B. $\frac{1}{1-x}$
• C. $\frac{1}{(1+x{)}^2}$
• D. $\frac{1}{(1-x{)}^3}$

Answer 1

The correct option is D $\frac{1}{(1-x{)}^3}$

Let $S=1+3x+{6x}^2+{10x}^3+...\infty \to \left(1\right)$
Multiplying $\left(1\right)$ by $x$ we get
$xS=x+{3x}^2+{6x}^3+...\to \left(2\right)$
$\left(1\right)-\left(2\right)$, gives
$(1-x)S=1+2x+{3x}^2+{4x}^2+...\to \left(3\right)$
Multiplying $\left(3\right)$ by $x$, we get
$x(1-x)S=x+{2x}^2+{3x}^3+{4x}^4+...\to \left(4\right)$
$\left(3\right)-\left(4\right)$, gives
${(1-x)}^{2}S=1+x+x^2+x^3+...$
$\Rightarrow {(1-x)}^{2}S=\frac{1}{1-x}\Rightarrow S=\frac{1}{{(1-x)}^3}$