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The sum of n terms of the series 1+2(1+1n)+3(1+1n)2+... is -

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Question

The sum of n terms of the series `1 + 2(1 + 1/n) + 3(1 + 1/n)^2 + ...` is

Options

  • `n^2`

  • `n(n + 1)`

  • `n(1 + 1/n)^2`

  • `(1 + 1/n)^2`

MCQ

Solution

`n^2`

Explanation:

Given series is `1 + 2(1 + 1/n) + 3(1 + 1/n)^2 + ......`

nth term = `n(1 + 1/n)^(n - 1)`

∴ Sum of n terms = `sumn(1 + 1/n)^(n - 1) = sumn(1 + (n - 1)/n)`

By using Binomial expansion and neglecting the remaining terms

= `sumn((n + n - 1)/n)`

= `sum 2n - 1`

= `(2(n + 1))/2 - n`

= `n^2 + n - n`

= `n^2`

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