..., 98, 99, 100!

For any real number $a$ , we have $\frac{1}{a(a+1)} = \frac{1}{a} - \frac{1}{a+1}$ . Thus: $S = \frac{1}{90} - \frac{1}{91} + \frac{1}{91} + ... + \frac{1}{99} - \frac{1}{100}$ . This telescopes to give us $S = \frac{1}{90} - \frac{1}{100} = \frac{1}{900}$ which gives the answer of $1 + 900 = 901$

$S=\left(\frac{1}{90}-\frac{1}{91}\right)+\left(\frac{1}{91}-\frac{1}{92}\right)+\cdots +\left(\frac{1}{99}-\frac{1}{100}\right)$ or, $S=\left(\frac{1}{90}-\frac{1}{100}\right)=\frac{1}{900}$ Then the answer follows.

A fraction with a product in the denominator can be split into two fractions in subtraction. For example: \­( \frac {1}{a \times b} ) where \­( b > a ) can be divided into \­( \frac {1}{a} - \frac{1}{b}. ) Similarly \­( \frac {1}{90 \times 91} ) can be split into \­( \frac {1}{90} - \frac {1}{91} ). and so can \­( {1}{91 \times 92} ) to \­( \frac {1}{91} - \frac{1}{92} ). But the second partial from the previous term and the first partial fraction from the second term cancel each other out, in the case of the first two terms \­( \frac {1}{91} ) is cancelled out. This goes on until \­( \frac {1}{99) ) gets cancelled and the only remaining terms are \­( \frac {1}{90} ) and $- \frac {1}{100}$ in addition.

That is equal to \­( \frac {10}{9000} ) i.e. \­( \frac {1}{900} ) and it is in the form \­( \frac {a}{b} ) where \­( a = 1 ) and \­( b = 900 ). Hence \­( a + b = 901 ).

=\frac {2}{90 92} + \frac {2}{92 94} + \frac {2}{94 96} + \frac{2}{96 98} + \frac {2}{98*100}

=\frac {4}{90 94} + \frac {4}{94 98} + \frac {2}{98*100}

=\frac {8}{90 98} + \frac {2}{98 100}

=\frac {1}{900}

partial the 1/90*91=1/r-1/r+1 then using summation to find the answer 1/90-1/91+1/91-1/92+....-1/100 =1/90-1/100 =100-90/9000=1/900 S=a/b a+b=901

That is same as $\frac {1}{90}-\frac {1}{91}+\frac {1}{91}-\frac {1}{92}+....\frac{1}{99}-\frac {1}{100}=\frac {1}{90}-\frac {1}{100}=\frac {1}{900}$

Notice that $\frac {1}{n \times (n+1)} = \frac {n+1 - n} { n \times (n+1)} = \frac {1}{n} - \frac {1}{n+1}$ . We use the method of difference to obtain $\begin{aligned} S &= \frac {1}{90} - \frac {1}{91} + \frac {1}{91} - \frac {1}{92} + \frac {1}{92} - \frac {1}{93} + \ldots + \frac {1}{98} - \frac {1}{99} + \frac {1}{99} - \frac {1}{100} \\ &= \frac {1}{90} - \frac {1}{100} \\ &= \frac {100 - 90 }{90 \times 100} \\ &= \frac {1}{900} \\ \end{aligned}$

Hence $a+b = 1 + 900 = 901$ .

S=190×91+191×92+192×93+193×94+194×95+195×96+196×97+197×98+198×99+199×100 S=1/900 BUT S=a/b so a/b=1/900 thus a+b=1+900=901