I have a problem that is supposed to use induction, but I have no idea how to solve it. Could I get some help?
The closed form sum of
12 $\left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]$ for $n \geq 1$ is $n(n+1)(n+2)(an+b)$. Find an + b.
On
Another way to do it is prove (which you probably already have) inductively the formula for the sum of squares and cubes: $$ 1^{2}+\ldots+n^{2} = \frac{n(n+1)(2n+1)}{6}\\ 1^{3}+\ldots+n^{3} = \frac{n^{2}(n+1)^{2}}{4} $$
and then note that the sum that you have $(1^{2}2+\ldots+n^{2}(n+1))$ is actually the sum of the first n squares and n cubes. Then, you have:
$$ 12(1^{2}2+\ldots+n^{2}(n+1)) = 12(\frac{n(n+1)(2n+1)}{6} + \frac{n^{2}(n+1)^{2}}{4}) \\ =n(n+1)(3n(n+1)+2(2n+1)) = n(n+1)(3n^{2}+7n+2) $$
So you're missing the factor of $n+2$. To finish then, you just do
$$ (n+2)(an+b)=(3n^{2}+7n+2), $$
open it up and realize that $a=3$ and $b=1$.
On
$$12\bigl(1^2\cdot 2+2^2\cdot 3+\dots+n^2(n+1)\bigr)=n(n+1)(n+2)(an+b)$$ for $n=1$ $$12\bigl(1^2\cdot 2\bigr)=1(2)(3)(a+b)=6(a+b)=6a+6b$$ $$24=6a+6b$$ $$a+b=4$$ $$b=4-a$$ for $n=2$ $$12\bigl(1^2\cdot 2+2^2\cdot 3\bigr)=2(3)(4)(2a+b)=24(2a+b)=48a+24b$$ $$12\cdot 14=48a+24b$$ $$14=4a+2b$$ $$7=2a+b$$ $$7=2a+4-a=a+4\implies a=3,b=4-3=1$$
For determining $a$ and $b$ it's sufficient to use two values of $n$, say $n=1$ and $n=2$. Then verifying the conjecture, that is $$ 12\bigl(1^2\cdot 2+2^2\cdot 3+\dots+n^2(n+1)\bigr)=n(n+1)(n+2)(an+b) $$ for all $n\ge 1$ (where $a$ and $b$ are those determined before) requires induction. You already have the base case ($n=1$), so you can write \begin{multline} 12\bigl(1^2\cdot 2+2^2\cdot 3+\dots+n^2(n+1)+(n+1)^2(n+2)\bigr)\\ =n(n+1)(n+2)(an+b) +12(n+1)^2(n+2) \end{multline} and go on from here to obtain the expression $$ (n+1)(n+2)(n+3)(an+a+b) $$