Sum of the series P=121+2+132+23+…+110099+99100 is

Let tr denote the rth term of the series, then tr=1(r+1)r+rr+1=1r(r+1)1r+1+r =r+1−rr(r+1)=1r−1r+1Thus, P=∑r=199 1r−1r+1=1−1100=910

Sum of the series P=121+2+132+23+…+110099+99100 is

Let tr denote the rth term of the series, then tr=1(r+1)r+rr+1=1r(r+1)1r+1+r =r+1−rr(r+1)=1r−1r+1Thus, P=∑r=199 1r−1r+1=1−1100=910

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

Sum of the series $P = \frac{1}{2 \sqrt{1} + \sqrt{2}} + \frac{1}{3 \sqrt{2} + 2 \sqrt{3}} + \dots + \frac{1}{100 \sqrt{99} + 99 \sqrt{100}}$ is

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

$9 / 10$

$3 / 10$

$1 / 2$

$1 / 10$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $t_{r}$ denote the $r t h$ term of the series, then
$\begin{array}{l} t_{r} = \frac{1}{(r + 1) \sqrt{r} + r \sqrt{r + 1}} = \frac{1}{\sqrt{r (r + 1)}} \frac{1}{\sqrt{r + 1} + \sqrt{r}} \\ = \frac{\sqrt{r + 1} - \sqrt{r}}{\sqrt{r (r + 1)}} = \frac{1}{\sqrt{r}} - \frac{1}{\sqrt{r + 1}} \end{array}$
Thus, $P = \sum_{r = 1}^{99} [\frac{1}{\sqrt{r}} - \frac{1}{\sqrt{r + 1}}] = 1 - \frac{1}{\sqrt{100}} = \frac{9}{10}$