Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

How to find this summation $\sum ^{999}_{n=1}\dfrac {1}{\sqrt [3] {n^{2}+2n+1}+\sqrt [3] {n^{2}+n}+\sqrt [3] {n^{2}}}$

share|cite|improve this question

HINT

Notice that $a^3-b^3= (a-b)(a^2+ab+b^2)$.

This gives $$ \dfrac {1}{\sqrt [3] {n^{2}+2n+1}+\sqrt [3] {n^{2}+n}+\sqrt [3] {n^{2}}}=\frac{\sqrt [3]{n+1}-\sqrt [3]n}{n+1 -n} $$

share|cite|improve this answer
    
Is it telescopic?? The answer is 9 right? – user109004 Dec 11 '13 at 10:12
    
Yes. That's right. – i707107 Dec 11 '13 at 10:13
    
Thanks for your hint ^^ – user109004 Dec 11 '13 at 10:19

Your Answer

 
discard

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.