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Question: The value of \(\sum_{r = 1}^{n}{( - 1)^{r - 1}\left( 1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}...

The value of ∑r=1n(−1)r−1(1+12+13+....+1r)\sum_{r = 1}^{n}{( - 1)^{r - 1}\left( 1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{r} \right)}nCr is equal to

A

–1

B

1r\frac{1}{r}

C

1n\frac{1}{n}

D

−1n3- \frac{1}{n^{3}}

Answer

1n\frac{1}{n}

Explanation

Solution

∑(−1)r−1.nCr(11+12+13+....+1r)\sum_{}^{}{( - 1)^{r - 1}.^{n}C_{r}\left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{r} \right)}

= ∑((−1)r−1.nCr∫01(1+x+x2+...+xr−1)dx)\sum_{}^{}\left( ( - 1)^{r - 1}.^{n}C_{r}\int_{0}^{1}{(1 + x + x^{2} + ... + x^{r - 1})dx} \right)

= ∑(−1)r−1.nCr∫01(1−xr1−x)dx\sum_{}^{}{( - 1)^{r - 1}.^{n}C_{r}\int_{0}^{1}\left( \frac{1 - x^{r}}{1 - x} \right)}dx

= ∫01∑r=1n(−1)r−1.nCr−(−1)r−1.nCrxr1−xdx\int_{0}^{1}{\sum_{r = 1}^{n}{\frac{( - 1)^{r - 1}.^{n}C_{r} - ( - 1)^{r - 1}.^{n}C_{r}x^{r}}{1 - x}dx}}

= ∫01−C0+1+(1−x)n1−xdx\int_{0}^{1}{\frac{- C_{0} + 1 + (1 - x)^{n}}{1 - x}dx}= ∫01(1−x)n−1dx\int_{0}^{1}{(1 - x)^{n - 1}dx}

= 1n\frac{1}{n}