$\prod_{k=1}^{n} \Big( k^2 + k \Big)$

Question

Consider this product:

$$\beta_{n} = \prod_{k=1}^{n} \Big( k^2 + k \Big)$$

I have tried applying associativity rule:

$$\beta_{n} = \prod_{k=1}^{n} \Big(k^2 + k\Big) = \prod_{k=1}^{n}k^2 + \prod_{k=1}^{n}k = (n!)^2 + n! = n! (n! + 1)$$

For $n = 4$ this produces output $600$. But it should be $2880$.

WolframAlpha gives this as a result: $\beta_{n} = (n!)^2(n+1)$

Hence I assume that associativity rule does not work like it would work with sums.

So how do I go about computing this quite simple product?

Answer 1

As @Winther mentioned in the comments, "a product of a sum is not the sum of a product": You tried to use a formula of the form $\prod_{k=1}^n(a_k+b_k)=\prod_{k=1}^na_k+\prod_{k=1}^n b_k$, but this is not true even for $n=2$: $$\prod_{k=1}^2(a_k+b_k)=a_1a_2+a_1b_2+b_1a_2+b_1b_2,\quad\text{whereas}\quad \prod_{k=1}^2a_k+\prod_{k=1}^2 b_k=a_1a_2+b_1b_2.$$

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In general, for sequences $(a_k)_k$ and $(b_k)_k$, we have \begin{align*} \prod_{k=1}^n a_kb_k &=(a_1b_1)(a_2b_2)\cdots(a_nb_n)\\ &=(a_1a_2\cdots a_n)(b_1b_2\cdots b_n)\\ &=\left(\prod_{k=1}^n a_k\right)\left(\prod_{k=1}^n b_k\right) \end{align*}

We apply this for $a_k=k$, $b_k=k+1$: \begin{align*} \prod_{k=1}^n(k^2+k)&=\prod_{k=1}^nk(k+1)\\ &=\left(\prod_{k=1}^nk\right)\left(\prod_{k=1}^n(k+1)\right)\\ &=n!(n+1)!\\ &=(n!)^2(n+1) \end{align*}

Answer 2

\begin{align*} &\prod_{k=1}^{n} \bigl(k^2 + k \bigr)\\[4pt] =\;&\prod_{k=1}^{n} \bigl(k(k+1)\bigr)\\[4pt] =\;&\left(\prod_{k=1}^{n} k\right)\!\!\left(\prod_{k=1}^{n} (k+1)\right)\\[4pt] =\;&n!{\,\cdot\,}(n+1)!\\[4pt] =\;&n!{\,\cdot\,}\bigl((n+1){\,\cdot\,}n!\bigr)\\[4pt] =\;&(n+1)(n!)^2\\[4pt] \end{align*} As regards your attempt . . .

To see that your attempted simplification $$\prod_{k=1}^{n} \Big(k^2 + k\Big) = \prod_{k=1}^{n}k^2 + \prod_{k=1}^{n}k$$ is not valid, note that for $n=2$, the LHS evaluates to $$(1^2+1)(2^2+2)=(2)(6)=12$$ whereas the RHS evaluates to $$(1^2)(2^2)+(1)(2)=4+2=6$$

Answer 3

You can't distribute the product: $$\prod_{k=1}^{n}(a_k+b_k)\ne \prod_{k=1}^na_k+\prod_{k=1}^n b_k.$$ However, $$\prod_{k=1}^n(a_kb_k)=\left(\prod_{k=1}^n a_k\right)\left(\prod_{k=1}^n b_k\right).$$ That being said, we have $$\begin{align} \prod_{k=1}^nk(k+1)&=\left(\prod_{k=1}^n k\right)\left(\prod_{k=1}^n (k+1)\right)\\ &=n!\prod_{k=2}^{n+1} k\\ &=n!(n+1)\prod_{k=2}^{n} k\\ &=n!^2(n+1) \end{align}$$