I'm really not sure how to solve these type of questions could somebody lead me in the right direction?
$$(a)\quad u_n = 3n^2+4n+5 \quad for \quad n=0,1,2,... $$ $$(b) \quad u_n = {n+7 \choose4} \quad for \quad n=0,1,2,...$$
The trick is usually to use the formula $$ \sum_{n=0}^\infty \binom{n+k}{k} x^n = \frac{1}{(1-x)^{k+1}}. $$ You can express $3n^2+4n+5$ as a linear combination of $\binom{n+2}{2},\binom{n+1}{1},\binom{n+0}{0}$, thereby solving part (a).
You can solve part (b) in a similar way, but there is a shortcut. We know what $\sum_{n=0}^\infty \binom{n+4}{4} x^n$ equals to. Notice that $$ \sum_{n=0}^\infty \binom{n+7}{4} x^n = \frac{1}{x^3} \sum_{n=0}^\infty \binom{n+7}{4} x^{n+3} = \frac{1}{x^3} \sum_{n=3}^\infty \binom{n+4}{4} x^n = \frac{1}{x^3} \left[ \sum_{n=0}^\infty \binom{n+4}{4} x^n - \sum_{n=0}^2 \binom{n+4}{4} x^n \right]. $$
The first is the generating function $$ \sum_{n=0}^\infty(3\,n^2+4\,n+5)t^n = 3\sum_{n=0}^\infty n^2\,t^n+4\sum_{n=0}^\infty n\,t^n+5\sum_{n=0}^\infty t^n\tag{1} $$ But a bit of calculus and the identity $$ \sum_{n=0}^\infty t^n=\frac{1}{1-t}\tag{2} $$ gives us \begin{align*} \sum_{n=0}^\infty n^2\,t^n&=\frac{t(t+1)}{(1-t)^3} & \sum_{n=0}^\infty n\,t^n&=\frac{t}{(1-t)^2}\tag{3} \end{align*} Can you use (2) to prove (3)? Can you plug the identities from (3) into (1) to obtain the answer?
