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 $a_n$ :

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



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.