Question

# The sum of infinity of the series $$\displaystyle 1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+$$..... is

A
1635
B
118
C
3516
D
86

Solution

## The correct option is C $$\displaystyle\frac{35}{16}$$Let $$\displaystyle S=1+\frac { 4 }{ 5 } +\frac { 7 }{ { 5 }^{ 2 } } +\frac { 10 }{ { 5 }^{ 3 } } +...$$   ...(1)Multiply (1) by $$\displaystyle \frac { 1 }{ 5 }$$ , we get$$\displaystyle \frac { 1 }{ 5 } S=\frac { 1 }{ 5 } +\frac { 4 }{ { 5 }^{ 2 } } +\frac { 7 }{ { 5 }^{ 3 } } +\frac { 10 }{ { 5 }^{ 4 } } +...$$   ...(2)$$(1) -(2)$$, gives $$\displaystyle \left( 1-\frac { 1 }{ 5 } \right) S=1+\frac { 3 }{ 5 } +\frac { 3 }{ { 5 }^{ 2 } } +\frac { 3 }{ { 5 }^{ 3 } } +...$$$$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 5 } \left( 1+\frac { 1 }{ 5 } +\frac { 1 }{ { 5 }^{ 2 } } +... \right)$$$$\displaystyle \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 1 }{ 1-\dfrac { 1 }{ 5 } } \right) \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 5 }{ 4 } \right)$$$$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 4 } \Rightarrow S=\frac { 35 }{ 16 }$$Mathematics

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