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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`A. `n f (n )-1 `B. `(n + 1 ) f(n) - n`C. `(n+ 1 ) f(n) +n `D. ` n f(n) +n ` |
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Answer» Correct Answer - B 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` |
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