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Question

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


A
45
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B
55
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C
Sum of first nine natural numbers
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D
Sum of first ten natural numbers
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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 $$\displaystyle 1+2+3+..9=\frac { 9\left( 9+1 \right)  }{ 2 } =45$$

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