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Question Number 85857 by M±th+et£s last updated on 25/Mar/20
Answered by mind is power last updated on 26/Mar/20
n2=14((2n−1)2+2(2n−1)+1))∑+∞n=1(−1)n−1n2(2n−1)3=∑n⩾1(−1)n−14(2n−1)+12∑n⩾1(−1)n−1(2n−1)2+14∑n⩾1(−1)n−1(2n−1)3∑n⩾1(−1)n−1(2n−1)=∑n⩾0(−1)n(2n+1)=S1tan−1(x)=Σ(−1)nx2n+12n+1S1=tan−1(1)=π4S2=∑n⩾1(−1)n−1(2n−1)2=∑n⩾0(−1)n(2n+1)2arctan(x)=Σ(−1)nx2n+12n+1⇒∫0ttan−1(x)xdx=∫0tΣ(−1)nx2n(2n+1)dx=∑n⩾0(−1)nt2n+1(2n+1)2⇒S2=∫01tan−1(x)xdxBypartS2=[ln(x)tan−1(x)]01−∫01ln(x)1+x2dx=−∫01ln(x)x2+1dx=GS3=∑n⩾0(−1)n(2n+1)3=∫011t∫0ttan−1(x)xdx=[ln(t)∫0ttan−1(x)xdx]01−∫01ln(t)tan−1(t)tdt=−∫01ln(t)ttan−1(t)dt=−[tan−1(t)ln2(t)2]+12∫01ln2(t)1+t2dt=S3ln(t)=limx→1∂∂x.tx−112.limy→1∂2∂y2∫01ty−11+t2dt=t2=z12limy→1∂2∂y2∫01zy2−11+z.dz2=14limy→1∂2∂y2∫01zy2−11+zdz14limy→1∂2∂y2.πsin(yπ2)=limx→1.∂∂x.−π2cos(πx2)8sin2(πx2)=π316=S3∑n⩾1(−1)n−1n2(2n−1)3=S14+S22+S34=π16+G2+π364=1128(8π+64G+2π3)
Commented by M±th+et£s last updated on 26/Mar/20
godblessyousir.nicesolution
Commented by mind is power last updated on 26/Mar/20
thanxSirhopgoodblessevreyone
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