Question

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. None of these

Answer 1

Answer: (D)

Let $x=1.1$, then
$S=1+2x+3x^2+.....+10x^9...\left(i\right)$
$Sx=x+2x^2+.....+9x^2+10x^{10}...\left(ii\right)$
Subtracting $\left(ii\right)$ from $\left(i\right)$, we get
$S\left(1+x\right)=1+x+x^2+.....+x^9-10x^{10}$
$=\frac{x^{10}-1}{x-1}-10x^{10}$
$=\frac{x^{10}-1}{0.1}-10x^{10}$, since $x=1.1$
$=10\left(x^{10}-1\right)-10x^{10}$
$=-10$
$\therefore S=\frac{10}{x-1}$
$=\frac{10}{0.1}$
$=100$.