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Question Number 26751 by abdo imad last updated on 28/Dec/17
byusingfourierseriefindthevalueof∑n=0∝1(2n+1)2
Commented by abdo imad last updated on 30/Dec/17
weknowthat/x/=π2−4π∑p=0∝cos(2p+1)x(2p+1)2(thefuctionis2πperiodiceven)⇒lettakex=00=π2−4π∑p=0∝1(2p+1)2⇒∑p=0∝1(2p+1)2=π28.
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