# Thread: Help again on mathematical induction. Also checking an answer as well.

1. ## Help again on mathematical induction. Also checking an answer as well.

First, here's the answer I want to check.

$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}=\frac{n}{2n+1}$
$s_1=\frac{1}{3} (check)$
$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}+\frac{1}{(2k+1)(2k+3)}$
$=P_k+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+1}$

~~~~~~~~~~~~~~~~~~~

Second, on the problem:

$2+7+...+(5n-3)=\frac{n}{2}(5n-1)$

I get to:

$P_k+(5k+2)$
$\frac{k}{2}(5k+1)+(5k+2)$

And once again, I get stuck trying to finish the problem off. I know how to get to this point, I just fail to see how to finish it off

Any and all help (and verification for the first problem) will be greatly appreciated.

2. Originally Posted by mathgeek777
First, here's the answer I want to check.

$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}=\frac{n}{2n+1}$
$s_1=\frac{1}{3} (check)$
$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1} {(2n-1)(2n+1)}+\frac{1}{(2k+1)(2k+3)}$
$=P_k+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+1}$

~~~~~~~~~~~~~~~~~~~

Second, on the problem:

$2+7+...+(5n-3)=\frac{n}{2}(5n-1)$

I get to:

$P_k+(5k+2)$
$\frac{k}{2}(5k+1)+(5k+2)$

And once again, I get stuck trying to finish the problem off. I know how to get to this point, I just fail to see how to finish it off

Any and all help (and verification for the first problem) will be greatly appreciated.
For the first problem, you did the induction correctly, except note that $\frac{k(2k + 3)}{2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3}$, and that is the answer you are looking for. I assume you just made a typo.

For the second problem, first note that the expression to be simplified is $\frac{k}{2}(5k - 1) + 5k + 2$, not $\frac{k}{2}(5k+1)+(5k+2)$. Try simplifying it as $\frac{1}{2}(5k^2 - k + 10k + 4)$ and factor.

3. Originally Posted by icemanfan
For the first problem, you did the induction correctly, except note that $\frac{k(2k + 3)}{2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3}$, and that is the answer you are looking for. I assume you just made a typo.
No typo. That's how I worked it out.

For the second problem, first note that the expression to be simplified is $\frac{k}{2}(5k - 1) + 5k + 2$, not $\frac{k}{2}(5k+1)+(5k+2)$. Try simplifying it as $\frac{1}{2}(5k^2 - k + 10k + 4)$ and factor.
Thanks for the assistance!