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
$\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