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)

$45$
sum of first nine natural numbers

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

First term

${\sum }_{r=1}^{\infty }2r{\left|\frac{9}{11}\right|}^r{S}_1=2(1.{\left|\frac{9}{11}\right|}^1+2.{\left|\frac{9}{11}\right|}^2+3.{\left|\frac{9}{11}\right|}^3+......\infty )\frac{9}{11}{S}_1=2\left(\right(\frac{9}{11}{)}^2+2.{\left(\frac{9}{11}\right)}^3+3.{\left(\frac{9}{11}\right)}^4+.....\infty )S_1-\frac{9}{11}{S}_1=2(\frac{9}{11}+{\left(\frac{9}{11}\right)}^2+{\left(\frac{9}{11}\right)}^3+........\infty )\frac{2S_1}{11}=2(\frac{9}{11}+{\left(\frac{9}{11}\right)}^2+{\left(\frac{9}{11}\right)}^3+........\infty )\frac{S_1}{11}={\sum }_{r=1}^{\infty }{\left(\frac{9}{11}\right)}^{r}So,S_1-{\sum }_{r=1}^{\infty }{\left|\frac{9}{11}\right|}^r=11{\sum }_{r=1}^{\infty }{\left|\frac{9}{11}\right|}^r-{\sum }_{r=1}^{\infty }{\left|\frac{9}{11}\right|}^r=10{\sum }_{r=1}^{\infty }{\left|\frac{9}{11}\right|}^r-(i)$

${\sum }_{r=1}^{\infty }{\left|\frac{9}{11}\right|}^r$ is in $GP$ and sum to infinity , common difference $=\frac{9}{11}$ and first term $=\frac{9}{11}$.

As diference is less than $1$.

So,Sum$=\frac{\frac{a}{11}}{1-\frac{a}{11}}$

$=\frac{a}{2}$

Putting the value in $\left(i\right)$

$=10\times \frac{a}{2}=45$

Answer $\left(A\right)$