Let `f(n)=1+1/2+1/3++1/ndot` Then `f(1)+f(2)+f(3)++f(n)` is equal to `nf(n)-1` (b) `(n+1)f(n)-nn` `(n+1)f(n)+n` (d) `nf(n)+n`

Options

A. `n f (n )-1 `
B. `(n + 1 ) f(n) - n`
C. `(n+ 1 ) f(n) +n `
D. ` n f(n) +n `

Answer

B

Explanation

We have, `f(n) = 1+(1)/(2)+(1)/(3) +……+(1)/(n)`
`therefore f(1) + f(2) + …….+f(n)`
`= 1+(1+(1)/(2)) +(1+(1)/(2)+(1)/(3)) +.....+(1+(1)/(2)+(1)/(3)+.....+(1)/(n))`
`= n +((n-1))/(2)+((n-2))/(3) +....+(n-(n-1))/(n)`
`=n(1+(1)/(2)+(1)/(3)+....+(1)/(n))-((1)/(2) +(2)/(3)+.....+(n-1)/(n))`
`=n(1+(1)/(2)+(1)/(3)+.....+(1)/(n))-{(1-(1)/(2))+(1-(1)/(3))+....+(1-(1)/(n))}`
`=n (1+(1)/(2)+(1)/(3)+.....+(1)/(n))-{(n-1)-((1)/(2)+(1)/(3)+....+(1)/(n))}`
`= n f (n) - {(n-1)-(f(n) -1)}`
`= (n+1) f(n) -n`