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I devised this question based on recent (and not-so-recent) MSE questions on summations.

Evaluate $$\frac{1\cdot2+2\cdot3+3\cdot4+\cdots+1000\cdot1001}{1\cdot1000+2\cdot999+3\cdot998+\cdots+1000\cdot1}$$

or more formally

Evaluate $$\frac {\displaystyle\sum_{r=1}^nr\ (r+1)\quad} {\displaystyle\sum_{r=1}^nr\ (n+1-r)} $$ for $n=1000$.

The obvious approach would be to evaluate the summation in the numerator and denominator separately and take the ratio in the hope that terms cancel out to give a nice answer.

Given that there is a nice answer, would it be possible to solve this using any other approach?

Hypergeometricx
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  • This is a normal A.P. question ... solve them independently ... no tricks involved! – NeilRoy Jun 01 '15 at 14:49
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    The limit is $$\frac{\int_0^1x^2\mathrm dx}{\int_0^1x(1-x)\mathrm dx}=\frac{\frac13}{\frac12-\frac13}=2.$$ Once this is noted, prove by an easy induction on $n$ that $$3\sum_{r=1}^{n}r^2=(2n+1)\sum_{r=1}^{n}r$$ to conclude that every ratio is actually $$2.$$ – Did Jun 01 '15 at 14:54
  • @Did - Thanks - that's a very good solution! The substitution really helps as we can cancel $\displaystyle\sum_{r=1}^nr$ without evaluating it first. The limit using integral is quite novel. If you would care to post this as a solution I would upvote and accept it. – Hypergeometricx Jun 01 '15 at 15:25
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    @NeilRoy - This is slightly more complicated than a normal AP... :) – Hypergeometricx Jun 01 '15 at 15:26

2 Answers2

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Do the sums for 2 or 3 instead of 1000; then 4,5,... and spot a pattern.

Empy2
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$$\sum_{r=1}^n r(n+1-r)=\sum_{r=1}^n\sum_{k=r}^n r=\sum_{1\le r\le k\le n}r=\sum_{k=1}^n\sum_{r=1}^kr=\sum_{k=1}^n{k+1\choose 2}={n+2\choose 3}$$ and $$\sum_{r=1}^n r(r+1)=2\sum_{r=1}^n{r+1\choose 2}=2{n+2\choose 3}$$

hxthanh
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