Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for that I require only odd powers $t^{2n-1}$, but I have no idea how to get them or what to do with the even part. Can anybody give me a hint how to start?
1 The expansion is $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$

\begin{align*}
  \sqrt{t} \sinh \sqrt{t} &= \sum_{n=1}^{\infty} \frac{t^{n}}{(2n-1)!} \\
  (\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n-1}}{(2n-1)!} \\
  t(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n}}{(2n-1)!} \\
  [t(\sqrt{t} \sinh \sqrt{t})']' &=
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n-1}}{(2n-1)!} \\
  t[t(\sqrt{t} \sinh \sqrt{t})']' &=
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} \\
  \sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} &=
  t\left[t
     \left(
       \frac{\sinh \sqrt{t}}{2\sqrt{t}}+
       \sqrt{t} \cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}
     \right)
   \right]' \\
  &=t\left(
       \frac{\sqrt{t}}{2}\sinh \sqrt{t}+\frac{t}{2} \cosh \sqrt{t}
     \right)' \\
  &=t\left(
       \frac{1}{4\sqrt{t}}\sinh \sqrt{t}+
       \frac{\sqrt{t}}{2}\cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}+
       \frac{1}{2} \cosh \sqrt{t}+
       \frac{t}{2} \sinh \sqrt{t} \times \frac{1}{2\sqrt{t}}
     \right) \\
  &= \frac{(t+1)\sqrt{t}}{4} \sinh \sqrt{t}+
     \frac{3t}{4} \cosh \sqrt{t}
\end{align*}
