Question

If $y=1+2x+3x^2+4x^3+\cdots +\infty$, where $|x|<1$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is C $\frac{2}{(1-x{)}^3}$

$y=1+2x+3x^2+4x^3+\cdots +\infty \to${i}

Multiplying by $x$ on both sides,
$yx=0+x+2x^2+3x^3+\cdots +\infty \to${ii}

Subtracting {ii} from {i}, we get
$y(1-x)=1+x+x^2+x^3+\cdots +\infty$

Using formula for infinite G.P
$y(1-x)=\frac{1}{(1-x)}$

$y=\frac{1}{(1-x{)}^2}$

$\frac{dy}{dx}=\frac{2}{(1-x{)}^3}$