Why does $\frac{1+t}{(1-t)^3}=\sum_{n=0}^{\infty}(1+n)^2t^n$ Background: This is a step from a longer proof/exercise that  $\sum_{n=1}^{\infty}\tau (n)^2/n^s=\zeta(s)^4/\zeta(2s)$ for $\sigma>1$
Expanding the sum and using counting I get: 
$\frac{1+t}{(1-t)^3}=(1+t)(1+{3\choose 1}t+\left ({3\choose 1}+{3\choose 2}\right )t+\left({3 \choose1}+3!+1\right )t^2 \ldots=\sum_{n=0}^{\infty}(1+n)^2t^n$$
So the final coefficients are each coefficient in the infinite sum plus the coefficient before it because of the $(1+t)$
This gives 1, 4, 9, 16
How would one calculate explicitly (as the author likely did) that the coefficients of $\frac{1+t}{(1-t)^3}$ are $(n+1)^2$ rather than counting and recognizing a pattern?
 A: Hint. One may start with the standard evaluation,
$$
\sum_{n=0}^\infty t^{n+1}=\frac{t}{1-t}, \qquad |t|<1. \tag1
$$ Then one is allowed to differentiate $(1)$ termwise and by multiplying by $t$ one gets
$$
\sum_{n=0}^\infty (n+1)t^{n+1}=\frac{t}{(1-t)^2}, \qquad |t|<1, \tag2
$$ one is allowed to differentiate $(2)$ termwise getting
$$
\sum_{n=0}^\infty(n+1)^2t^n=\frac{1+t}{(1-t)^3}, \qquad |t|<1, \tag3
$$ as announced.
A: Using Negative binomial theorem:
$$\frac{1+t}{(1-t)^3}=(1+t)(1-t)^{-3}=(1+t)\sum_{k=0}^{\infty}{-3\choose k}(-t)^k=\\
\color{blue}{(1+t)\sum_{k=0}^{\infty}{2+k\choose k}t^k}=\sum_{k=0}^{\infty}\left[\color{red}{{2+k\choose k}+{1+k\choose k-1}}\right]t^k=
\sum_{k=0}^\infty (1+k)^2t^k.$$
Note:

$$\color{blue}{(1+t)\sum_{k=0}^{\infty}{2+k\choose k}t^k}=\sum_{k=0}^\infty{2+k\choose k}t^k+\sum_{k=0}^\infty{2+k\choose k}t^{k+1}=\\\sum_{k=0}^\infty{2+k\choose k}t^k+\sum_{k=0}^\infty{1+k\choose k-1}t^k\\\sum_{k=0}^{\infty}\left[{2+k\choose k}+{1+k\choose k-1}\right]t^k.\\\color{red}{{2+k\choose k}+{1+k\choose k-1}}=\\{2+k\choose 2}+{1+k\choose 2}=\frac12(2+k)(1+k)+\frac12(1+k)k=(1+k)^2.$$

