edited by
8,044 views
44 votes
44 votes
$\sum\limits_{x=1}^{99}\frac{1}{x(x+1)}$ = ______.
edited by

5 Answers

Best answer
58 votes
58 votes
$\frac{1}{1.2}+\frac{1}{2.3}+ \dots +\frac{1}{99.100}\\=1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}+ \dots +\frac{1}{99} - \frac{1}{100}\\=1-\frac{1}{100} = 0.99$
selected by
81 votes
81 votes
It can be answered by using the concept of partial fraction

$\dfrac{1}{x\left(x+1\right)} =\dfrac{A}{x} + \dfrac{B}{\left(x+1\right)}$

solving this we will get $A=1$ and $B=-1$

So, this will form a sequence in which $2$ terms will remain $\left(1-\dfrac{1}{100}\right)$ and we will get $\dfrac{99}{100}=0.99$ as answer.
edited by
28 votes
28 votes
If u take 2 instead of 99 u will get 2/3

If u take 3  u will get 3/4

Means behavior remains same...

So for 99 u will get 99/100
4 votes
4 votes
$\sum_{x = 1}^{99} = 1/x *(1+x) = (1/x) - (1/1+x)$

$(1/x) - (1/1+x) = (1+x - x) / x*(x+1) = 1/x *(1+x) $

$s_{1} = 1 - 1/2$

$s_{2} = 1/2 - 1/3$

$s_{3} = 1/3 - 1/4 $

$.$

$.$

$.$

$.$

$.$

$s_{99} = 1/99 - 1/100$

on adding all, we get

$s_{1} + s_{2} + s_{3} + ...............................................+ s_{99}= \sum_{x = 1}^{99} = 1 - 1/100 = 99/100 = 0.99$
Answer:

Related questions

80 votes
80 votes
7 answers
1
makhdoom ghaya asked Feb 13, 2015
28,736 views
The least number of temporary variables required to create a three-address code in static single assignment form for the expression $q + r / 3 + s - t * 5 + u * v/w$ is_...
32 votes
32 votes
9 answers
2
makhdoom ghaya asked Feb 13, 2015
24,370 views
Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is_______________.