Question

find the sum 1/1+√2 +1/√2+√3 + 1/√3+√4...... UPTO 99 terms

Answer 1

Given:

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+..........99\;terms$

$=\frac{1}{\sqrt1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+..........99\;terms$

consider,

$\frac{1}{\sqrt1+\sqrt2}$

$=\frac{1}{\sqrt2+1}{\times}\frac{\sqrt2-1}{\sqrt2-1}$

$=\frac{\sqrt2-1}{(\sqrt2)^2-1^2}$

$=\frac{\sqrt2-1}{2-1}$

$=\sqrt2-1$

$\implies\;t_1=\frac{1}{\sqrt1+\sqrt2}=\sqrt2-1$

and

$\frac{1}{\sqrt2+\sqrt3}$

$=\frac{1}{\sqrt3+\sqrt2}{\times}\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}$

$=\frac{\sqrt3-\sqrt2}{(\sqrt3)^2-(\sqrt2)^2}$

$=\frac{\sqrt3-\sqrt2}{3-2}$

$=\sqrt3-\sqrt2$

$\implies\;t_2=\frac{1}{\sqrt2+\sqrt3}=\sqrt3-\sqrt2$

k th term of the series is

$\bf\;t_k=\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$

Now,

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+..........99\;terms$

$=(\sqrt{2}-\sqrt1)+(\sqrt{3}-\sqrt2)+(\sqrt{4}-\sqrt3)+...................+(\sqrt{99}-\sqrt{98})+(\sqrt{100}-\sqrt{99})$

$=-\sqrt1+\sqrt{100}$

$=-1+10$

$=9$

$\implies\boxed{\bf\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+..........99\;terms=9}$

Answer 2

Answer:

9.

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