Use induction to prove the following:
$1a)$ Show for all positive integers $n$ that $$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$
$1b)$ Show that if $p$ and $q$ are polynomials so that
$i)$ $q(0)=0$
$ii)$ $p(n)=q(n)-q(n-1)$ for all $n$ that
$$ \sum_{i=1}^n p(i) = q(n) $$ for all positive integers $n$.
$1c)$ Show that for any polynomial $p$, there exists a polynomial $q$ so that $$ \sum_{i=1}^n p(i) = q(n) $$ for all positive integers $n$. [Hint: use induction on the degree of $p$ and note that $n^d-(n-1)^d = dn^{d-1}+\textrm{lower order terms}.$]
I have for 1a)
Base Case: n=1
$$ \sum_{i=1}^1 i^2 = \frac{1(1+1)(2+1)}{6}=1 $$
Inductive Step: $$ \sum_{i=1}^1 i^2+(n+1)^2 = \frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{2n^3+9n^2+13n+6}{6}=\frac{(n+1)(n+2)(2n+3)}{6}=\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} $$
I'm having a hard time understanding 1b) and 1c). I'm not sure where to start.
For 1bii,
$$\sum_{i=1}^n p(i) =\sum_{i=1}^n (q(i)-q(i-1)) =(q(1)-q(0))+(q(2)-q(1))+...+(q(n)-q(n-1)) $$ and note that everything cancels out except $q(n)-q(0)$.
Note: I do not think that the problem should have assumed that $q(0) = 0$.