nth term of a certain sequence, a_n

Question

n^{th}  term of a certain sequence,  a_n=\frac{1}{n^2}-\frac{1}{(n+1)^2} . What is the sum of first 100 terms of the given series?

(A)  1

(B)  \frac{1}{101^2}

(C)  \frac{102}{101^2}

(D)  \frac{10200}{101^2}

(E)  \frac{102}{50}

Aabhash777 · Answer

a1 = 1 - 1/4
a2 = 1/4 - 1/9
a3 = 1/9 - 1/16
.....
a100 = 1/(100^2) - 1/ (101)^2

If you take the sum of all the terms, the terms cancel each other out and we have

S100 = 1 - 1/(101)^2
 = 10,200 / 101^2

Answer D

hr1212 · Answer

Pattern for these type of questions is mostly similar ie. the middle terms will cancel each other out except the first and last term which is,

1 - 1/(101)^2

Now wait and think... Don't calculate the square here but look for better number manipulation.

If possible, try to optimize each step to save those extra few secs.

(101^2 - 1)/101^2

a^2-b^2 = (a-b)*(a+b)

(101-1)*(101+1) = 100*102 = 10200

= 10200/101^2

IMO: D

Carcass · Answer

Alright, let's tackle this problem step by step. I'm going to find the sum of the first 100 terms of the sequence defined by:

$$
 a_n=\frac{1}{n^2}-\frac{1}{(n+1)^2} 
$$

Understanding the Sequence
First, let's write out the first few terms to see if we can spot a pattern.
- $ a_1=\frac{1}{1^2}-\frac{1}{2^2}=1-\frac{1}{4} $
- $ a_2=\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9} $
- $ a_3=\frac{1}{3^2}-\frac{1}{4^2}=\frac{1}{9}-\frac{1}{16} $
- ...
- $ a_{100}=\frac{1}{100^2}-\frac{1}{101^2}=\frac{1}{10000}-\frac{1}{10201} $

Summing the Terms
Now, let's consider the sum $S$ of the first 100 terms:

$$
 S=a_1+a_2+a_3+\cdots+a_{100} 
$$

Substituting the expressions for each $a_n$ :

$$
 S=\left(\frac{1}{1^2}-\frac{1}{2^2}ight)+\left(\frac{1}{2^2}-\frac{1}{3^2}ight)+\left(\frac{1}{3^2}-\frac{1}{4^2}ight)+\cdots+\left(\frac{1}{100^2}-\frac{1}{101^2}ight) 
$$

Observing the Pattern
When we write out the sum like this, we can see that many terms cancel out:

$$
 S=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots+\frac{1}{100^2}-\frac{1}{101^2} 
$$

Notice that $ -\frac{1}{2^2}+\frac{1}{2^2}=0,-\frac{1}{3^2}+\frac{1}{3^2}=0 $, and so on. This is called a telescoping series, where most terms cancel out when we sum them.

After cancellation, we're left with:

$$
 \begin{aligned}
S & =\frac{1}{1^2}-\frac{1}{101^2} \
S & =1-\frac{1}{101^2}
\end{aligned} 
$$

Calculating $ 101^2 $
Now, let's compute $ 101^2 $ :

$$
 101^2=10201 
$$

So,

$$
 S=1-\frac{1}{10201} 
$$

Simplifying the Expression
To combine these into a single fraction:

$$
 \begin{gathered}
S=\frac{10201}{10201}-\frac{1}{10201} \
S=\frac{10200}{10201}
\end{gathered} 
$$

But looking back at the options, none directly match $ \frac{10200}{10201} $. Wait, let's see the options again:
(A) 1
(B) $ \frac{1}{101^2} $
(C) $ \frac{102}{101^2} $
(D) $ \frac{10200}{101^2} $
(E) $ \frac{102}{50} $

Ah, I see! $ 101^2=10201 $, so $ \frac{10200}{10201}=\frac{10200}{101^2} $, which is option D.
Wait, but is $ \frac{10200}{101^2} $ equal to $ \frac{10200}{10201} $ ? Yes, because $ 101^2=10201 $.
But earlier, I thought the sum was $ \frac{10200}{10201} $, which is $ \frac{10200}{101^2} $, so option D seems correct.
But let me double-check the sum:

$$
 S=1-\frac{1}{101^2}=\frac{101^2-1}{101^2}=\frac{10201-1}{10201}=\frac{10200}{10201}=\frac{10200}{101^2} 
$$

Yes, that's correct. So option D matches our result.

nth term of a certain sequence, a_n

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

nth term of a certain sequence, a_n

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards