Closed Form Summations

Question

Problem:

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 $an + b.$

How would I start this? Is there an easy way?

Answer 1

Let $f(n) = 12 \cdot [1^2 \cdot 2 + 2^2 \cdot 3 + \cdots + n^2 \cdot (n + 1)]$. We have $$ f(1) = 24, \quad f(2) = 168 $$ Therefore, $$ 6(a + b) = 24, \quad 24(2a + b) = 168 $$ That is, $$ a + b = 4, \quad 2a + b = 7 $$ Solving these equations, we get $$ a = 3,\quad b = 1 $$

Answer 2

Hint :

$$ \sum n^3=\Bigg(\sum n\Bigg)^2=\frac{n^2(n+1)^2}{4}$$

$$ \sum n^2=\frac{n(n+1)(2n+1)}{6}$$

And your sum is :

$$12 \sum n^2(n+1)= 12 \Big(\sum n^3+\sum n^2 \Big)$$

Answer 3

One may note by approximating with an integral that

$$\sum_{k=1}^nk^2(k+1)=\frac14n^4+\mathcal O(n^3)$$

Thus $a=3$ and by subsituting $n=1$, we find $b=1$.

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Alternatively, One may notice that

$$\sum_{k=1}^nk^2(k+1)=\sum_{k=1}^nk^3+k^2=\frac{6n^4+20n^3+18n^2+4n}{24}$$

which follows from Faulhaber's formula.

Answer 4

$n^2(n+1)=n(n+1)(n+2)-2n(n+1)=n(n+1)(n+2)\frac{n+3-(n-1)}{4}-2n(n+1)\frac{n+2-(n-1)}{3}$ $=\frac{1}{4}\Delta[n(n+1)(n+2)(n+3)]-\frac{2}{3}\Delta[n(n+1)(n+2)]$

$\displaystyle\sum_{k=1}^n k^2(k+1)=\frac{1}{4}n(n+1)(n+2)(n+3)-\frac{2}{3}n(n+1)(n+2)=n(n+1)(n+2)\left(\frac{n+3}{4}-\frac{2}{3}\right)=\frac{n(n+1)(n+2)(3n+1)}{12}$