advanced-mathematics-prove-that-n-0-1-2-n-3n-n-3-125-11pi-6-2log-2-45- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 127725 by mnjuly1970 last updated on 01/Jan/21 …advancedmathematics…provethat::∑∞n=012n(3nn)=???3125(11π6−2log(2)+45) Answered by Ar Brandon last updated on 01/Jan/21 1+2!2⋅3!+4!2!22⋅6!+6!3!239!+⋅⋅⋅(2n)!n!2n(3n)!+⋅⋅⋅Sn=∑∞n=0Γ(2n+1)Γ(n+1)2nΓ(3n+1)=∑∞n=0nΓ(2n+1)Γ(n)2nΓ(3n+1)=∑∞n=0n2nβ(2n+1,n)=∑∞n=0n2n∫01x2n(1−x)n−1dx=∫01{∑∞n=0[n2nx2n(1−x)n−1]}dx=∫01{11−x⋅∑∞n=0n(x2(1−x)2)n}dxS(t)=∑∞k=0tk=11−t⇒S′(t)=∑∞k=0ktk−1=1(1−t)2tS′(t)=∑∞k=0ktk=t(1−t)2∑∞n=0n(x2(1−x)2)n=x2(1−x)2(1−x2(1−x)2)2=x2(1−x)2⋅4(2−x2+x3)2=2x2(1−x)(x3−x2+2)2Sn=∫01{11−x⋅2x2(1−x)(x3−x2+2)2}dx=∫012x2dx(x3−x2+2)2=∫012x2((x+1)(x2−2x+2))2f(x)=2(x(x+1)(x2−2x+2))2=225(x+2x2−2x+2−1x+1)2=225(x2+4x+4(x2−2x+2)2−2(x+2)(x+1)(x2−2x+2)+1(x+1)2)=225(1x2−2x+2+6x+2(x2−2x+2)2−25(x+1)+2(x−1)x2−2x+2+1(x+1)2)6x+2=3(2x−2)+8∫01f(x)dx=225[tan−1(x−1)−3x2−2x+2+8∫dx((x−1)2+1)2[−2ln(x+1)5−ln(x2−2x+2)−1x+1]01Sn=225(−3−2ln25−12)−225(−π4−32−ln2−1)−2π25=−225+6ln2125−6π100Typos Commented by Dwaipayan Shikari last updated on 01/Jan/21 ∑∞n=012n(2nn)=∑∞n=0Γ2(n+1)2nΓ(2n+1)=∫01∑∞n=0n2nxn(1−x)n−1dx=2∫01x(2−x(1−x))2dx=∫012x−1(x2−x+2)2+∫011(x2−x+2)2dx=∫011((x−1+i72)(x−1−i72))2dx=−17∫011(x−1+i72)2+1(x−1−i72)2−2(x2−x+2)=17∫011x2−x+2dx+17[.2x−1x2−x+2]01=877tan−117+17 Commented by mnjuly1970 last updated on 01/Jan/21 nicevetynicr Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: e-3x-2-1-x-4-1-8-dx-Next Next post: nice-calculus-evaluate-0-pi-2-0-pi-2-ln-cos-x-2-ln-cos-y-2-cos-x-cos-y-dxdy- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.