expression for$\sum_{k=0}^{n}(k+1)(k+2)$

Question

what is a closed expression for$\sum_{k=0}^{n}(k+1)(k+2)$? the answer says to look at Generating function $a_k=(k+1)(k+2)$ and derive twice $\dfrac{1}{x}=1+x+x^2+...$ so the solution starts: $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$but isn't $\sum_{k=0}^{\infty}t^k=\dfrac{1}{(1-t)}$? so how the lhs and rhs in $\sum_{k=0}^{n}(k+1)(k+2)=[\dfrac{1}{(1-x)}.\dfrac{2}{(1-x)^3}]$ are equal?

Answer 1

Hint. You may want to write

$$ \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} (k+1)(k+2) \right) x^{n} = \frac{2}{(1-x)^{4}}. $$

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Here is another solution using the Pascal's triangle:

$$ \sum_{k=0}^{n} (k+1)(k+2) = 2 \sum_{k=0}^{n} \binom{k+2}{2} = 2 \left[ \binom{2}{2} + \cdots + \binom{n+2}{2} \right]. $$

Now we know that $\binom{2}{2} = 1 = \binom{3}{3}$. So by applying the property $\binom{m}{k-1} + \binom{m}{k} = \binom{m+1}{k}$ repeatedly, we observe

\begin{align*} \binom{2}{2} + \cdots + \binom{n+2}{2} &= \binom{3}{3} + \binom{3}{2} + \binom{4}{2} + \cdots + \binom{n+2}{2} \\ &= \binom{4}{3} + \binom{4}{2} + \cdots + \binom{n+2}{2} \\ &= \binom{5}{3} + \cdots + \binom{n+2}{2} \\ &\quad \quad \vdots \\ &= \binom{n+2}{3} + \binom{n+2}{2} \\ &= \binom{n+3}{3}. \end{align*}

So we obtain a special case of Christmas Stocking Theorem:

$$ \sum_{k=0}^{n} (k+1)(k+2) = 2 \sum_{k=0}^{n} \binom{k+2}{2} = 2 \binom{n+3}{3} = \frac{(n+3)(n+2)(n+1)}{3}. $$

Answer 2

Isn't it like this:

$$ \sum_{k=0}^n (k+1)(k+2) = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2+k = \sum_{k=1}^n k^2 + \sum_{k=1}^n k = \sum_{k=1}^n k^2 + \frac{n(n+1)}{2} $$

Answer 3

Using the generalization proven in this answer, and the substitutions $$ \begin{align} M&\mapsto n\\ m&\mapsto k\\ k&\mapsto 2\\ n&\mapsto 0 \end{align} $$ we get $$ \begin{align} \sum_{k=0}^n(k+1)(k+2) &=2\sum_{k=0}^n\binom{k+2}{2}\binom{n-k}{0}\\ &=2\binom{n+3}{3} \end{align} $$

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An Alternate Approach

The recursive definition of the Pascal Triangle says that $$ \binom{k+2}{2}=\binom{k+3}{3}-\binom{k+2}{3} $$ Therefore, $$ \begin{align} \sum_{k=0}^n(k+1)(k+2) &=\sum_{k=0}^n2\binom{k+2}{2}\\ &=2\left[\sum_{k=0}^n\binom{k+3}{3}-\sum_{k=0}^n\binom{k+2}{3}\right]\\ &=2\left[\sum_{k=0}^n\binom{k+3}{3}-\sum_{k=-1}^{n-1}\binom{k+3}{3}\right]\\ &=2\left[\binom{n+3}{3}-\binom{2}{3}\right]\\ &=2\binom{n+3}{3} \end{align} $$

Answer 4

$$ \sum_{k=0}^n (k+1)(k+2) = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n k^2+k=$$ $$=\sum_{k=1}^n k^2+\sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}=$$ $$=\frac{n(n+1)}{2}\left(\frac{2n+1}{3}+1\right)=\frac{n(n+1)}{2}\frac{2n+4}{3}=\frac{n(n+1)(n+2)}{3}$$

Answer 5

By induction it can easily be shown that: $$\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$$ This for $r=0,1,2,\dots$

For $r=2$ this results in: $$\frac{1}{2}\sum_{k=2}^{n}\left(k-1\right)k=\frac{1}{6}\left(n-1\right)n\left(n+1\right)$$ or equivalently: $$\sum_{k=0}^{n}\left(k+1\right)\left(k+2\right)=\frac{1}{3}\left(n+1\right)\left(n+2\right)\left(n+3\right)$$