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Question

Question: Sum of the series S = 1 +\(\frac{1}{2}\) (1 + 2) +\(\frac{1}{3}\) (1 + 2 + 3) + \(\frac{1}{4}\) (1 +...

Sum of the series S = 1 +12\frac{1}{2} (1 + 2) +13\frac{1}{3} (1 + 2 + 3) + 14\frac{1}{4} (1 + 2 + 3 + 4) …….. up to 20 terms is

A

110

B

111

C

115

D

116

Answer

115

Explanation

Solution

tr = 1r\frac{1}{r} (1 + 2 + 3 ….. + r) = 1r.12\frac{1}{r}.\frac{1}{2}r (r + 1)

S = åtr = 12r=1n(r+1)\frac{1}{2}\sum_{r = 1}^{n}{(r + 1)} = 115