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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
Sum of the series P=121+2+132+23+…+110099+99100P=\frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\ldots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}} is
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Correct option: C
9/109 / 10
Option C (9/10) is correct because the series telescopes, and after simplification, the sum equals 9/10.

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