Question

# $$\displaystyle \sum_{r\, =\, 1 }^{\infty}\, (2r\, -\, 1)\, \left ( \frac{9}{11} \right )^r$$ is equal to

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

Solution

## The correct options are A 45 C Sum of first nine natural numbers$$\displaystyle S=\sum _{ r=1 }^{ \infty }{ \left( 2r-1 \right) { \left( \frac { 9 }{ 11 } \right) }^{ r } }$$$$\displaystyle S=\frac { 9 }{ 11 } +3{ \left( \frac { 9 }{ 11 } \right) }^{ 2 }+5{ \left( \frac { 9 }{ 11 } \right) }^{ 3 }+....$$   ...(1)Multiplying (1) by $$\displaystyle \frac { 9 }{ 11 }$$, we get$$\displaystyle \frac { 9 }{ 11 } S={ \left( \frac { 9 }{ 11 } \right) }^{ 2 }+3{ \left( \frac { 9 }{ 11 } \right) }^{ 3 }+5{ \left( \frac { 9 }{ 11 } \right) }^{ 4 }+...$$   ...(2)(1) - (2) gives $$\displaystyle \left( 1-\frac { 9 }{ 11 } \right) S=\frac { 9 }{ 11 } +2{ \left( \frac { 9 }{ 11 } \right) }^{ 2 }+2{ \left( \frac { 9 }{ 11 } \right) }^{ 3 }+...$$$$\displaystyle \Rightarrow \frac { 2S }{ 11 } =\frac { 9 }{ 11 } +2{ \left( \frac { 9 }{ 11 } \right) }^{ 2 }\left( \frac { 1 }{ \left( 1-\frac { 9 }{ 11 } \right) } \right)$$$$\displaystyle \Rightarrow \frac { 2S }{ 11 } =\frac { 9 }{ 11 } +2{ \left( \frac { 9 }{ 11 } \right) }^{ 2 }\left( \frac { 11 }{ 2 } \right)$$$$\displaystyle \Rightarrow \frac { 2S }{ 11 } =\frac { 9 }{ 11 } +\frac { 81 }{ 11 } =\frac { 90 }{ 11 }$$$$\displaystyle \Rightarrow S=\frac { 90 }{ 2 } =45$$And $$\displaystyle 1+2+3+..9=\frac { 9\left( 9+1 \right) }{ 2 } =45$$Maths

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