The sum of $n$ terms of the series $1+\dfrac { 4 }{ 5 } +\dfrac { 7 }{ { 5 }^{ 2 } } +\dfrac { 10 }{ { 5 }^{ 3 } } +\cdots $ is
- $\dfrac { 5 }{ 4 } +\dfrac { 15 }{ 16 } \left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { \left( 3n-2 \right) }{ 4\cdot { 5 }^{ n-1 } } $
- $\dfrac { 5 }{ 4 } +\dfrac { 1 }{ 16 } \left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { 3n }{ 4\cdot { 5 }^{ n-1 } } $
- $\left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { \left( 3n+2 \right) }{ 4\cdot { 5 }^{ n-1 } } $
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None of the above
This is an arithmetico-geometric series. The sum of the first n terms follows the standard formula for such series.
This is an arithmetico-geometric progression where the numerator forms an arithmetic progression 1, 4, 7, 10 with the nth term (3n - 2), and the denominator forms a geometric progression 1, 5, 25, 125 with the nth term 5^(n-1). To find the sum S, use the standard method of multiplying the series by the common ratio of the GP (5) and subtracting the result from the original series. This eliminates the numerators, leaving you with a simple geometric series and a single final term. Solving for S gives 5/4 + 15/16(1 - 1/5^(n-1)) - (3n - 2)/(4*5^(n-1)).