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Question

# If S∞=11⋅3⋅5+13⋅5⋅7+15⋅7⋅9+⋯ up to infinite term, then the value of 24S∞ is

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Solution

## Tr=1(2r−1)(2r+1)(2r+3)=14[1(2r−1)(2r+1)−1(2r+1)(2r+3)]⇒Sn=14n∑r=1[1(2r−1)(2r+1)−1(2r+1)(2r+3)]⇒4Sn=(11⋅3−13⋅5) +(13⋅5−15⋅7) +(15⋅7−17⋅9) ⋮ +(1(2n−1)(2n+1)−1(2n+1)(2n+3))⇒4Sn=13−1(2n+1)(2n+3)⇒4S∞=13∴24S∞=2

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