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 $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$

Let's carry this computation out for instructional purposes.
Let
$$f(t) = \sum_{n=1}^{\infty} \frac{t^n}{(2 n)!} = \cosh{\left (\sqrt{t}\right )}-1$$
Then the sum we seek is
$$2 \sum_{n=1}^{\infty} \frac{n^3 t^n}{(2 n)!} = \left (t \frac{d}{dt} \right )^3 f(t) = 2 t f'(t) + 6 t^2 f''(t) + 2 t^3 f'''(t)$$
$$f'(t) = \frac{\sinh{\left (\sqrt{t}\right )}}{2 \sqrt{t}} $$
$$f''(t) = -\frac14 t^{-3/2} \sinh{\left (\sqrt{t}\right )} + \frac14 t^{-1}  \cosh{\left (\sqrt{t}\right )}$$
$$f'''(t) = \frac18 t^{-5/2} (3+t) \sinh{\left (\sqrt{t}\right )} - \frac38 t^{-2} \cosh{\left (\sqrt{t}\right )} $$
Putting this all together, I get that the sum equals

$$\sum_{n=1}^{\infty} \frac{n^2 t^n}{(2 n-1)!} = \frac14 t (1+t) \frac{\sinh{\left (\sqrt{t}\right )}}{\sqrt{t}} + \frac34 t \cosh{\left (\sqrt{t}\right )} $$

