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

This question was asked in (selection) IMO for 8th graders.

$1/2 + 1/6 + 1/12+ 1/20 + 1/30 + 1/42 +1/56 + 1/72 + 1/90 + 1/110 +1/132$

I have noticed that it can be written as $1/(1*2) + 1/(2*3) +1/(3*4) + 1/(4*5).... +1/(11*12)$

However I don't know how to continue..

share|cite|improve this question
    
Parentheses, please. $1/1*2$ can be read as $(1/1)*2$ which is clearly not what you mean. – Ross Millikan Feb 13 at 2:35
up vote 4 down vote accepted

Hint:

Your sum can be written in the following form:

$$\sum\limits_{n=1}^{11} \dfrac{1}{n(n+1)}$$

Apply partial fraction decomposition to $\frac{1}{n(n+1)}$ to write it in the form $\frac{A}{n}+\frac{B}{n+1}$ and see what happens as you simplify.

The series should telescope. I.e. adjacent terms will cancel with one another, leaving you with only the first and last terms not cancelled. You have $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. The series then looks like $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\cdots-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}$

share|cite|improve this answer

Then note that $\frac 1{1\cdot 2}$ = $\frac 11 - \frac 12$ and the sum telescopes.

share|cite|improve this answer

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.