Question

The sum to $10$ terms of the series $1+2\left(1.1\right)+3{\left(1.1\right)}^2+4{\left(1.1\right)}^3+....$ is

• A. $85.12$
• B. $92.5$
• C. $96.75$
• D. $100$

Answer 1

Answer: (D)

Let $x=1.1$
$S=1+2x+3x^2+........10x^9$ and
$xS=x+2x^2+........9x^9+10x^{10}$
Subtracting, we get,
$S\left(1-x\right)=\left(1+x+x^2+.......+x^9\right)-10x^{10}$
$=\left(\frac{x^{10}-1}{x-1}\right)-10x^{10}$
$\Rightarrow -S=\frac{x^{10}-1}{{\left(x-1\right)}^2}-\frac{10x^{10}}{\left(x-1\right)}$
$\Rightarrow -S=\frac{{\left(1.1\right)}^{10}-1}{{\left(0.1\right)}^2}-\frac{10{\left(1.1\right)}^{10}}{0.1}=100\times {\left(1.1\right)}^{10}-100-100\times {\left(1.1\right)}^{10}=-100$
$\Rightarrow S=100.$