I have found this formula to a sum of multiples: $$\sum^a_{n=1}\prod^{b-1}_{x=0}(n+x)=\frac{\prod^{b}_{x=0}(a+x)}{b+1}$$For example, when $a=5$ and $b=3$, the sum would be $1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7$, and the formula would give $\frac{5\cdot6\cdot7\cdot8}{4}=420$. It seems to work for many values of $a$ and $b$, but how do I prove this formula?
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Do you mean $\sum^a_{n=1}\prod^{b-1}_{x=0}(n+x)$? – actinidia Dec 17 '17 at 04:33
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@TiwaAina Yes, I will make according changes. – Kyan Cheung Dec 17 '17 at 05:19
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@Kyky: Tiwa Aina says $0\le x\le b-1$ instead of $1\le x\le b$. – mathlove Dec 17 '17 at 05:22
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As Tiwa Aina suggests, the formula $$\sum^a_{n=1}\prod^b_{x=1}(n+x)=\frac{\prod^b_{y=0}a+y}{b+1}$$ is wrong.
This answer proves that $$\sum^a_{n=1}\prod^{\color{red}{b-1}}_{x=\color{red}{0}}(n+x)=\frac{\prod^b_{y=0}(a+y)}{b+1}$$
Considering binomial coefficients should help.
$$\begin{align}\sum_{n=1}^{a}\prod_{x=0}^{b-1}(n+x)&=\sum_{n=1}^{a}\frac{(n+b-1)!}{(n-1)!}\\\\&=b!\sum_{n=1}^{a}\frac{(n+b-1)!}{b!(n-1)!}\\\\&=b!\sum_{n=1}^{a}\binom{n+b-1}{b}\\\\&\color{red}{=}b!\binom{b+a}{b+1}\\\\&=\frac{b!(b+a)!}{(b+1)!(a-1)!}\\\\&=\frac{1}{b+1}\prod_{y=0}^{b}(a+y)\end{align}$$
The proof for the equality in red can be seen here.
mathlove
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