A Place for Latest Exam wise Questions, Videos, Previous Year Papers, Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
2
The value of the sum $\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}$ is
Solution
Step 1: General term
$T_n = \dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$
Step 2: Rationalize by multiplying numerator and denominator by $(n+1)\sqrt{n}-n\sqrt{n+1}$
Denominator becomes:
$[(n+1)\sqrt{n}]^2 - [n\sqrt{n+1}]^2$
$ = n(n+1)^2 - n^2(n+1) $
$= n(n+1)(n+1-n)$
$ = n(n+1)$
So:
$T_n = \dfrac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)}$
$= \dfrac{\sqrt{n}}{n} - \dfrac{\sqrt{n+1}}{n+1}$
$= \dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}$
Step 3: Telescoping sum from $n=1$ to $n=24$
$\displaystyle\sum_{n=1}^{24} T_n = \left(\dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}}\right) + \left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}}\right) + \cdots + \left(\dfrac{1}{\sqrt{24}} - \dfrac{1}{\sqrt{25}}\right)$
$= \dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{25}}$
$= 1 - \dfrac{1}{5}$
$= \dfrac{4}{5}$
Answer: $\boxed{\dfrac{4}{5}}$
Online Test Series, Information About Examination, Syllabus, Notification and More.