f(n)=1+12+13+….+1n then ∑i=1n f(i) is equal to

f(1)+f(2)+f(3)+……..+f(n)=1+1+12+1+12+13+…..+1+12+13+….+1n=n+n−12+n−23+…..+1n=n1+12+13+….+1n−12+23+…..+n−1n=nf(n)−1−12+1−13+…..+1−1n=nf(n)−(n−1)+f(n)−1=nf(n)−n+f(n)

f(n)=1+12+13+….+1n then ∑i=1n f(i) is equal to

f(1)+f(2)+f(3)+……..+f(n)=1+1+12+1+12+13+…..+1+12+13+….+1n=n+n−12+n−23+…..+1n=n1+12+13+….+1n−12+23+…..+n−1n=nf(n)−1−12+1−13+…..+1−1n=nf(n)−(n−1)+f(n)−1=nf(n)−n+f(n)

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$f (n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n} then \sum_{i = 1}^{n} f (i)$ is equal to

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$(n - 1) f (n)$

$n (f (n) - 1) + 1$

$n {f (n) - 1)} + f (n)$

$n {f (n) - 1)} - 1$

answer is C.

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Detailed Solution

$\begin{array}{l} f (1) + f (2) + f (3) + \dots \dots . . + f (n) \\ = 1 + (1 + \frac{1}{2}) + (1 + \frac{1}{2} + \frac{1}{3}) + \dots . . + (1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n}) = n + \frac{n - 1}{2} + \frac{n - 2}{3} + \dots . . + \frac{1}{n} \\ = n (1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n}) - (\frac{1}{2} + \frac{2}{3} + \dots . . + \frac{n - 1}{n}) \\ = nf (n) - \{(1 - \frac{1}{2}) + (1 - \frac{1}{3}) + \dots . . + (1 - \frac{1}{n})\} = nf (n) - (n - 1) + f (n) - 1 \\ = nf (n) - n + f (n) \end{array}$