Question

$\sum _{r=1}^{\infty }(2r-1){\left(\frac{9}{11}\right)}^r$ is equal to

• A. 45
• B. 55
• C. Sum of first nine natural numbers
• D. Sum of first ten natural numbers

Answer 1

Answer: (A), (C)

The correct options are
A 45
C Sum of first nine natural numbers

$S=\sum _{r=1}^{\infty }(2r-1){\left(\frac{9}{11}\right)}^r$
$S=\frac{9}{11}+3{\left(\frac{9}{11}\right)}^2+5{\left(\frac{9}{11}\right)}^3+....$ ...(1)
Multiplying (1) by $\frac{9}{11}$, we get
$\frac{9}{11}S={\left(\frac{9}{11}\right)}^2+3{\left(\frac{9}{11}\right)}^3+5{\left(\frac{9}{11}\right)}^4+...$ ...(2)
(1) - (2) gives
$(1-\frac{9}{11})S=\frac{9}{11}+2{\left(\frac{9}{11}\right)}^2+2{\left(\frac{9}{11}\right)}^3+...$
$\Rightarrow \frac{2S}{11}=\frac{9}{11}+2{\left(\frac{9}{11}\right)}^2\left(\frac{1}{(1-\frac{9}{11})}\right)$
$\Rightarrow \frac{2S}{11}=\frac{9}{11}+2{\left(\frac{9}{11}\right)}^2\left(\frac{11}{2}\right)$
$\Rightarrow \frac{2S}{11}=\frac{9}{11}+\frac{81}{11}=\frac{90}{11}$
$\Rightarrow S=\frac{90}{2}=45$
And $1+2+3+..9=\frac{9(9+1)}{2}=45$