# Mathematical Induction

• Jun 21st 2009, 11:41 PM
maybeline9216
Mathematical Induction
For qn 20a, can u show me how [1/(sqrrt2 +1)] is equal to (sqrt 2 -1)
by doing workings and nt using decimals.

And pls teach me qn21(i) the hence part.

Thks alot!!!
• Jun 22nd 2009, 04:40 AM
Amer
Quote:

Originally Posted by maybeline9216
For qn 20a, can u show me how [1/(sqrrt2 +1)] is equal to (sqrt 2 -1)
by doing workings and nt using decimals.

And pls teach me qn21(i) the hence part.

Thks alot!!!

ok we want to prove that

$\sum_{r=1}^{n} \frac{1}{\sqrt{r+1}+\sqrt{r}}=\sqrt{n+1}-1$

let
$P_k:\sum_{r=1}^{k}\frac{1}{\sqrt{r+1}+\sqrt{r}} = \sqrt{k+1}-1$

lats see if $p_k$ true or not

$P_1 :\frac{1}{\sqrt{2}+1} =\sqrt{2}-1$

$\left(\frac{1}{\sqrt{2}+1}\right)\left(\frac{\sqrt {2}-1}{\sqrt{2}-1}\right) = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$
so P1 is true suppose that Pk is true try P(k+1)

$P_{k+1} : \sum_{r=1}^{k+1}\frac{1}{\sqrt{r+1}+\sqrt{r}} = \sqrt{(k+1)+1}-1=\sqrt{k+2}-1$

the left side

$\sum_{r=1}^{k+1}\frac{1}{\sqrt{r+1}+\sqrt{r}}=\fra c{1}{\sqrt{k+2}+\sqrt{k+1}}+\sum_{r=1}^{k}\frac{1} {\sqrt{r+1}+\sqrt{r}}$

but the second term equal $\sqrt{k+1}-1$

$\frac{1}{\sqrt{k+2}+\sqrt{k+1}}+\sqrt{k+1}-1$

$\left(\frac{1}{\sqrt{k+2}+\sqrt{k+1}}\right)\left( \frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+2}-\sqrt{k+1}}\right)+\sqrt{k+1}-1$

$\Rightarrow \frac{\sqrt{k+2}-\sqrt{k+1}}{k+2-k-1}+\sqrt{k+1}-1$

$\sqrt{k+2}-\sqrt{k+1}+\sqrt{k+1}-1 =\sqrt{k+2}-1$

so $P_{k+1}$ is true we prove it by induction