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I am trying to prove something by induction, and in induction step I had to prove this $$1+ \sum_{k=1}^{\lceil{\frac{n-1}{2}}\rceil} (-1)^{k}\frac{(t^2)^{2k}}{(2k)!} = \sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor}(-1)^k \frac{(t^2)^{2k}}{(2k)!}. $$ Any Idea?

Edit : (the first one is solved) what about this equality $$ -\sum_{k=0}^{\lfloor{\frac{n-1}{2}}\rfloor}(-1)^{k-1} \frac{(t^2)^{2k+1}}{(2k+1)!} = \sum_{k=1}^{\lceil{\frac{n}{2}}\rceil} (-1)^{k-1}\frac{(t^2)^{2k-1}}{(2k-1)!} $$

J. Mccain
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You can embbed the "$1+$" as the 0 th summand in the left summatory, and then your equality follows from the fact that $\left\lceil\frac{n-1}{2}\right\rceil=\left\lfloor n/2\right\rfloor$. You can see this easily splitting the cases where $n$ is even and odd.

Edit:[You're very welcome] For the second equality just make the change of summation $k\leftarrow k-1$. Then the right sum is $$\sum_{k=0}^{\left\lceil n/2\right\rceil -1}(-1)^k\frac{(t^2)^{2k+1}}{(2k+1)!}=-\sum_{k=0}^{\left\lceil n/2\right\rceil -1}(-1)^{k-1}\frac{(t^2)^{2k+1}}{(2k+1)!}$$ And the equality follows cause $\left\lfloor\frac{n-1}{2}\right\rfloor=\left\lceil n/2\right\rceil-1$. You can check this again by splitting in the even and odd cases.

  • Thanks, that was very helpful. I added another equality to be checked in my post. – J. Mccain Jun 20 '16 at 00:22
  • Sorry, I found the solution for the second equation. Just change the index of the LHF with $k'=k+1$ where $k'$ is the new index. – J. Mccain Jun 20 '16 at 00:31
  • @J.Mccain I posted the second answer before O saw your comment. Gonna leave it like this. :) –  Jun 20 '16 at 00:41