# Please solve q 8?

##### 1 Answer
Feb 17, 2018

Answer is option $\left(4\right)$

#### Explanation:

General term, ${T}_{n} = \frac{1}{{\left(2 n - 1\right)}^{\frac{1}{2}} + {\left(2 n + 1\right)}^{\frac{1}{2}}}$

Rationalize the General term of the sequence :-

rArr T_n = ( (2n-1)^(1/2)-(2n+1)^(1/2)) /[( (2n-1)^(1/2)+(2n+1)^(1/2)).( (2n-1)^(1/2)-(2n+1)^(1/2))]

$\Rightarrow {T}_{n} = \frac{{\left(2 n - 1\right)}^{\frac{1}{2}} - {\left(2 n + 1\right)}^{\frac{1}{2}}}{- 2}$

$\Rightarrow {T}_{n} = \frac{{\left(2 n + 1\right)}^{\frac{1}{2}} - {\left(2 n - 1\right)}^{\frac{1}{2}}}{2}$

Now summing all the terms ie.

$\Rightarrow \sum {T}_{n} = {\sum}_{n = 0}^{n} \frac{{\left(2 n + 1\right)}^{\frac{1}{2}} - {\left(2 n - 1\right)}^{\frac{1}{2}}}{2}$

$\Rightarrow \sum {T}_{n} = \frac{\left(3 - 1\right) + \left(5 - 3\right) + \ldots \ldots \ldots \ldots . + \left({\left(2 n + 1\right)}^{\frac{1}{2}} - {\left(2 n - 1\right)}^{\frac{1}{2}}\right)}{2}$

$\Rightarrow \sum {T}_{n} = \frac{\left(\cancel{3} - 1\right) + \left(\cancel{5} - \cancel{3}\right) + \ldots \ldots \ldots \ldots . + \left({\left(2 n + 1\right)}^{\frac{1}{2}} - {\cancel{2 n - 1}}^{\frac{1}{2}}\right)}{2}$...........................($) $\therefore \sum {T}_{n} = \frac{{\left(2 n + 1\right)}^{\frac{1}{2}} - 1}{2}$....................option $\left(4\right)$$N O T E$:- In equation ($) we should note a Pattern of

cancellation of the terms has formed ; where the initial

terms are eliminated by the succeeding terms at some point ,

except the Least and the Biggest terms which are $1$ and

${\left(2 n + 1\right)}^{\frac{1}{2}}$ respectively and the term ${\left(2 n - 1\right)}^{\frac{1}{2}}$ also gets

cancelled.