If frac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+s | vedclass

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

Question

(D) For $m$: $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{k+1-k} = \sqrt{k+1}-\sqrt{k}$.Summing from $k=1$ to $99$: $(\sqrt{2}-\sqrt{1}

Vedclass · Accepted Answer

$11 x-100 y=0$

Vedclass · Answer

(D) For $m$: $\frac{1}{\sqrt{k}+\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{k+1-k} = \sqrt{k+1}-\sqrt{k}$.Summing from $k=1$ to $99$: $(\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + \ldots + (\sqrt{100}-\sqrt{99}) = \sqrt{100}-\sqrt{1} = 10-1 = 9$.So,$m=9$.For $n$: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.Summing from $k=1$ to $99$: $(1-\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + \ldots + (\frac{1}{99}-\frac{1}{100}) = 1-\frac{1}{100} = \frac{99}{100}$.So,$n=\frac{99}{100}$.The point is $(m, n) = (9, \frac{99}{100})$.Checking the options,for option $D$: $11(9) - 100(\frac{99}{100}) = 99 - 99 = 0$.Thus,the point lies on the line $11x-100y=0$.

If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$,then the point $(m, n)$ lies on the line

Similar Questions

The sum of the first $n$ terms of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ is

The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to

$\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \ldots$ to $50$ terms $=$

The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

If $\sum_{r=1}^{n} T_{r} = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$,then $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \left(\frac{1}{T_{r}}\right)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module