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}\quad...\left(i\right)$
$ Sx= x+2x^{2}+.....+9x^{2}+10x^{10} \quad...\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} -10\,x^{10} $
$=\frac{ x^{10}-1 }{0.1} -10\,x^{10}$, since $x=1.1$
$ = 10\left(x^{10}-1\right) -10\,x^{10} $
$= -10$
$\therefore S= \frac{10}{x-1}$
$ = \frac{10}{0.1} $
$=100$.