A question on proof by induction and cubic functions the full question is 

I equated each equation to f(n) and ended up with four equations and four unknowns. However I dont seem to be getting anywhere
Also does anyone know if this method will work for the series sum of r^3 ? 
any help is hugely appreciated! thank you :)
 A: Presumably you got the following system of equations:
$$\left\{\begin{align*}
&d=0\\
&a+b+c+d=1\\
&8a+4b+2c+d=5\\
&27a+9b+3c+d=14\;,
\end{align*}\right.$$
which immediately reduces to
$$\left\{\begin{align*}
&a+b+c=1\\
&8a+4b+2c=5\\
&27a+9b+3c=14\;.
\end{align*}\right.$$
Eliminate $c$ by subtracting multiples of the first equation from the other two:
$$\left\{\begin{align*}
&6a+2b=3\\
&24a+6b=11\;.
\end{align*}\right.$$
Subtracting the second equation from $4$ times the first yields $2b=1$, so $b=\frac12$, and $6a+1=3$, or $a=\frac13$. Finally, from $a+b+c=1$ we have $c=\frac16$, and the cubic polynomial is 
$$f(n)=\frac13n^3+\frac12n^2+\frac16n=\frac16\left(n\left(2n^2+3n+1\right)\right)=\frac16n(n+1)(2n+1)\;.$$
Now you have only to prove by induction that 
$$\sum_{k=0}^nk^2=\frac16n(n+1)(2n+1)$$
for $n\ge 0$.
Yes, this approach will work to find a closed form for $\sum_{k=0}^nk^m$ for any positive integer $a$, though it’s pretty tedious if $a$ is much bigger than $2$; $m=3$ is quite feasible, however.
