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Q:
Simplify arithmetic series with formula
I have this expression:
$3\left(\frac{n+1}{n}\right)^n+3\left(\frac{n+2}{n+1}\right)^n+\dots+3\left(\frac{n+n}{n}\right)^n$
I know that I can simplify the summation to:
$\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}+\dots+1=n^{n-1}+n^{n-2}+\dots+1=\frac{(n+1)(n-1)!}{n!}$
However, when I simplify the fraction in my summation, it seems to give me a different answer:
$\left(\frac{n+1}{n}\right)^n+\left(\frac{n+2}{n+1}\right)^n+\dots+\left(\frac{n+n}{n}\right)^n=\frac{(n+1)!}{n!}$
Why is this happening?
A:
If you have:
$\left(\frac{n+1}{n}\right)^n+\left(\frac{n+2}{n+1}\right)^n+\dots+\left(\frac{n+n}{n}\right)^n=\frac{(n+1)(n-1)!}{n!}$
then you have to have
$\frac{n+1}{n}+\frac{n+2}{n+1}+\dots+\frac{n+n}{n}=\frac{n!}{(n-1)!}$
or
$(n+1)+(n+1)+\dots+(n+1)=n!$
Hence $n=0$ so the series is indeed $3$.
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