chetan2u wrote:
What is the value of \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3....+50}\)?
(A) \(\frac{49}{100}\)
(B) \(\frac{49}{50}\)
(C) \(1\)
(D) \(\frac{49}{25}\)
(E) \(\frac{49}{10}\)
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Hi Chetan2u--I am getting 100/51
if we will see the denominator : it is sum of consecutive integer :
\(\frac{n*(n+1)}{2}\)
the expression can be broken down into
\(\frac{2}{n*n(n+1)}\) -- where n = 1 to 50
further we can write it as\(\frac{2}{n}\) - \(\frac{2}{n+1}\)----- where n = 1 to 50
now 1st term\(\frac{2}{1}\) -\(\frac{2}{2}\)---1st
\(\frac{2}{2}\) - \(\frac{2}{3}\)--2nd
..
similarly it will go till
\(\frac{2}{50}\) - \(\frac{2}{51}\)---50th
each fraction wil be cancel out except 2 -\(\frac{2}{51}\) =\(\frac{100}{51}\)