The sum of the series frac{1}{sqrt{1} + sqrt{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given series is $S = \sum_{k=1}^{n^2-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$.To solve this,we rationalize each term by multiplying the numerator a

Vedclass · Accepted Answer

$n - 1$

Vedclass · Answer

(D) The given series is $S = \sum_{k=1}^{n^2-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$.To solve this,we rationalize each term by multiplying the numerator and denominator by the conjugate $(\sqrt{k+1} - \sqrt{k})$:$\frac{1}{\sqrt{k+1} + \sqrt{k}} 	imes \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} = \frac{\sqrt{k+1} - \sqrt{k}}{(k+1) - k} = \sqrt{k+1} - \sqrt{k}$.Applying this to the series:$S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + ... + (\sqrt{n^2} - \sqrt{n^2 - 1})$.This is a telescoping series where intermediate terms cancel out:$S = -\sqrt{1} + \sqrt{n^2} = -1 + n = n - 1$.

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

Similar Questions

Find the sum of infinite terms of the series $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots$

The ratio of the $7^{th}$ to the $3^{rd}$ term of an $A.P.$ is $12:5$. Find the ratio of the $13^{th}$ to the $4^{th}$ term.

The sum of the first two terms of a $G.P.$ is $36$ and the product of the first and the third terms is $9$ times the second term,then find the sum of the first $8$ terms.

The $n^{th}$ term of the following series $(1 \times 3) + (3 \times 5) + (5 \times 7) + (7 \times 9) + \dots$ will be

Let $a_{n}$ be the $n^{\text{th}}$ term of a $G$.$P$. of positive terms. If $\sum_{n=1}^{100} a_{2n+1} = 200$ and $\sum_{n=1}^{100} a_{2n} = 100$,then $\sum_{n=1}^{200} a_{n}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module