Q.399. Sum of a finite series.

Question 399.
Find the sum of 1/(1*3) + 1/(3*5)+.....+1/(99*101).

Answer 399.
1/(1*3) + 1/(3*5) + ..... + 1/(99*101)
= (1/2) * [(3-1)/(1*3) + (5-3)/(5*3) + (7-5)/(7*5) + (9-7)/(9*7) + + (101-99)/(99*101)]
= (1/2) * [1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101]
= (1/2) * (1 - 1/101)
= (1/2) * (100)/(101)
= (50)/(101).

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