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!!!

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- Jun 21st 2009, 11:41 PMmaybeline9216Mathematical 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 AMAmer
ok we want to prove that

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

let

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

lats see if $\displaystyle p_k$ true or not

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

$\displaystyle \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)

$\displaystyle 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

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

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

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

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

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

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

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