Problem Solving

[Math Revolution GMAT math practice question]

If 1/n(n+1) = 1/n - 1/(n+1), then what is the value of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)

A. 1/100
B. 1/50
C. 49/50
D. 99/100
E. 1/2

Hi,

Given function, 1/n*(n+1) = 1/n - 1/(n+1),

For patterns and sequences questions all we need to do is find initial few values and we will get to understand the pattern to solve the question.

1/(1*2) = 1/1 - ½

1/(2*3) = ½ - 1/3

1/(3*4) = 1/3 - ¼

We can clearly see that, adding the first three terms in the given sum, we will get 1 - ¼ = ¾

Similarly when we add the entire series,

1/1 -1/2 + ½ - 1/3 +......-1/99 + 1/99 -1/00

We will get only first term minus the term left, as all the other terms will get cancelled(subtracted) off

So,

1/1 - 1/100 = 99/100

So the answer is D.

Hope this helps.

=>
1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100)
= (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100) = 1/1 - 1/100 = 1 - 1/100 = 99/100 after cancellation of the inner terms.
Therefore, the answer is D.
Answer: D