Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$

Question

The question

Sum the first $n$ terms of the series: $$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$

This was asked under the heading using method of difference and the answer given was $$ S_n = \frac{1}{10}n(n+1)(n+2)(n+3)(2n+3). $$

My approach

First, I get $$ U_n=n(n+2)(n+1)^2. $$

Then I tried to make $U_n = V_n - V_{n-1}$ in order to get $S_n = V_n - V_0$. But I really don't know how can I figure this out.

Answer 1

We have:

$$\sum_{r=2}^n{r^2(r+1)(r-1)}=\sum_{r=2}^n{r^2(r^2-1)}=\sum_{r=2}^n{r^4}-\sum_{r=2}^n{r^2}$$ Just use (or show) the well-known formulae to evaluate this.

Answer 2

\begin{align*} \sum_{k=1}^n k(k+2)(k+1)^2&=\sum_{k=1}^n (k+3)(k+2)(k+1)k-2\sum_{k=1}^n (k+2)(k+1)k\\ &=24 \sum_{k=1}^n \binom{k+3}{4}-12 \sum_{k=1}^n \binom{k+2}{3}\\ &=24\binom{n+4}{5}-12\binom{n+3}{4} \end{align*}

Answer 3

Rewrite the formula as $$\sum_2^{n+1}k^2(k+1)(k-1)=\sum_2^{n+1}k^4-\sum_2^{n+1}k^2.$$ It is a well known formula that $$1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$ Due to Fermat, we also have $$1+2^4+3^4+\cdots+n^4=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n.$$ Just substitute in your values to get the result.