Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find all positive integers n, such that $$\sum\limits_{k=1}^{n}(k+2)^n=(n+3)^n $$

share|cite|improve this question
Why don't you put the whole question in the title? – Jason DeVito Nov 1 '10 at 1:47
up vote 5 down vote accepted

$k^n/(n+3)^n = \left(1-\frac{n+3-k}{n+3}\right)^n$, so using that $\log (1+x) \leq x$ for all $x$, \[ \frac{\sum_{k=3}^{n+2} k^n}{(n+3)^n} \leq \sum_{j=1}^n e^{-jn/(n+3)} \leq \frac{1}{e^{n/(n+3)}-1} \] This last term is decreasing, and for $n=7$ it is strictly less than $1$, so you are left with only $n \leq 6$ to check. It is probably possible to sharpen the inequalities so that there would be less to check.

In the end, you get $n=2$ or $3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.