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Question Number 162424 by mnjuly1970 last updated on 29/Dec/21
calculateΩ=∑∞n=1(−1)nn3n(2n−1)=?−InspiredfromSirGhaderi′spost−
Answered by Ar Brandon last updated on 29/Dec/21
Ω=∑∞n=1(−1)nn3n(2n−1)=12∑∞n=1(−13)n+12∑∞n=1(−13)n12n−1=−18+12∑∞n=1(−13)n∫01x2n−2dx=−18+12∫011x2∑∞n=1(−x23)ndx=−18−12∫011x2⋅x231+x23dx=−18−12∫011x2+3dx=−18−[123arctan(x3)]01=−18−π123
Commented by mnjuly1970 last updated on 29/Dec/21
thankyousomuchsirbrandon
Commented by Ar Brandon last updated on 29/Dec/21
Mypleasure,Sir!
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