Can somebody teach me how to solve this question? Please…

Sorry... since i don't know how to write the symbols here... so i used attachment...

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- Oct 4th 2012, 02:41 AMAuXianSeries and sequences
Can somebody teach me how to solve this question? Please…

Sorry... since i don't know how to write the symbols here... so i used attachment... - Oct 4th 2012, 03:06 AMemakarovRe: Series and sequences
- Oct 4th 2012, 03:25 AMjohnsomeoneRe: Series and sequences
If it's $\displaystyle \sum_{r = 1}^n \frac{r^2 + r^{-1}}{r^2 + r} = \frac{n^2}{n+1}$, then it isn't even true.

$\displaystyle n =1:$ LHS $\displaystyle = \frac{1^2 +1^{-1}}{1^2 + 1} = \frac{2}{2} = 1$. RHS $\displaystyle = \frac{1^2}{1+1} = \frac{1}{2}$.

$\displaystyle n =2$: LHS $\displaystyle = (1) + \frac{2^2 +2^{-1}}{2^2 + 2} = 1 + \frac{4.5}{6} = \frac{7}{4}$. RHS $\displaystyle = \frac{2^2}{2+1} = \frac{4}{3}$. - Oct 4th 2012, 06:13 AMAuXianRe: Series and sequences
- Oct 4th 2012, 07:42 AMemakarovRe: Series and sequences
- Oct 4th 2012, 08:09 AMAuXianRe: Series and sequences
- Oct 4th 2012, 08:15 AMemakarovRe: Series and sequences
And yet you had to add the phrase "only the final step will convert r to n," the meaning of which I don't understand. Even worse, the left-hand side of the equation in the quote in post #2 does not depend on n (n is a bound variable used as the summation index) while the right hand side does. And if n and r are reversed, then, as post #3 shows, the equation is incorrect.