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 $$ -\sum_{k=0}^{\lfloor{\frac{n-1}{2}}\rfloor}(-1)^{k-1} \frac{(t^2)^{2k+1}}{(2k+1)!} = \sum_{k=1}^{\lceil{\frac{n}{2}}\rceil} (-1)^{k-1}\frac{(t^2)^{2k-1}}{(2k-1)!} $$
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.