Question

Prove by induction that n^2/(2n-1)(2n+1) = n(n+1)/2(2n+1)

Answer 1

\[\sum_{r=1}^{n} n ^{2}/ (2n-1)(2n+1) = n(n+1)/2(2n+1)\]

Answer 2

do you mean:\[\sum_{r=1}^{n} \frac{r^2}{(2r-1)(2r+1)} = \frac{n(n+1)}{2(2n+1)}\]

Answer 3

yes! im sorry for the weird looking equation

Answer 4

ok, so for induction, you first show it to be true for r=1, then assume it is true for r=k and use that to prove it is also true for r=k+1.

Answer 5

i understand the steps in induction although i cant seem to get it right for the part when proving r=k+1. when r=k+1, what do you equate?

Answer 6

ok, so proving it is true for r=1 is trivial. for r=k, we assume:\[\sum_{r=1}^{k} \frac{r^2}{(2r-1)(2r+1)} = \frac{k(k+1)}{2(2k+1)}\]we then need to compute:\[\sum_{r=1}^{k+1} \frac{r^2}{(2r-1)(2r+1)} = \sum_{r=1}^{k} \frac{r^2}{(2r-1)(2r+1)}+\frac{(k+1)^2}{(2(k+1)-1)(2(k+1)+1)}\]\[=\frac{k(k+1)}{2(2k+1)}+\frac{(k+1)^2}{(2(k+1)-1)(2(k+1)+1)}\]
you need to simplify this last expression to show that it is equal to:\[\frac{(k+1)((k+1)+1)}{2(2(k+1)+1)}\]
do you understand what to do from here on now?

Answer 7

hey thanks for the clear steps asnaseer, im still a bit confused on how you got the last expression the (k+1)((K+1)+1)/2(2(k+1)+1) though

Answer 8

it comes from replacing "k" with "k+1" in this expression:\[\sum_{r=1}^{k} \frac{r^2}{(2r-1)(2r+1)} = \frac{k(k+1)}{2(2k+1)}\]