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Question:

The sum to infinity of the series 1 + 2/3 + 6/32 + 10/33 + 14/34 + ... is:

2

4

6

3

Solution:

LetS=1+23+632+1033+1434+(1)Multiplying with13on both sides,S3=13+232+633+1034+1435+(2)∴(1)−(2)⇒2S3=1+13+432+433+434+...⇒2S3=4[13+132+133+134+...]⇒2S3=4⎡⎢⎢⎢⎣131𕒵3⎤⎥⎥⎥⎦[Sum of GP=a(1−rn)1−rwhenr<1]⇒2S3=4[12]∴S=3Hence, the answer is3.