If aK-1kk+1 for K=1, 2, 3......n, then∑K=1nak2=

ak=1k(k+1)=1k-1k+1∑k=1n ak=1-12+12-13+.....+1n-1n+1 =1-1n+1=nn+1∑k=1n akn=n2(n+1)2

If aK-1kk+1 for K=1, 2, 3......n, then∑K=1nak2=

ak=1k(k+1)=1k-1k+1∑k=1n ak=1-12+12-13+.....+1n-1n+1 =1-1n+1=nn+1∑k=1n akn=n2(n+1)2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If a_{K} - \frac{1}{k (k + 1)} for K = 1, 2, 3 . . . . . . n, then {[\sum_{K = 1}^{n} a_{k}]}^{2} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{n^{4}}{{(n + 1)}^{4}}$

$\frac{n^{6}}{{(n + 1)}^{6}}$

$\frac{n^{2}}{{(n + 1)}^{2}}$

$\frac{n}{n + 1}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} a_{k} & = & \frac{1}{k (k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \\ \sum_{k = 1}^{n} a_{k} & = & 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . . . + \frac{1}{n} - \frac{1}{n + 1} \\ = & 1 - \frac{1}{n + 1} = \frac{n}{n + 1} \\ {(\sum_{k = 1}^{n} a_{k})}^{n} & = & \frac{n^{2}}{{(n + 1)}^{2}} \end{array}$