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Question Number 78941 by M±th+et£s last updated on 21/Jan/20
Q.findthesumS=232!+333!+434!+....thenfind∑∞n=1n4n!
Answered by mind is power last updated on 21/Jan/20
=∑n⩾1n3(n−1)!=∑k⩾0(k+1)3k!=∑k⩾0(k3+3k2+3k+1)k!=∑k⩾1k2(k−1)!+3∑k⩾1k(k−1)!+3∑k⩾11(k−1)!+∑k⩾11k!=∑k⩾0(k+1)2k!+3∑k⩾0k+1k!+3∑k⩾01k!+∑k⩾01k!−1=∑k⩾0k2+2k+1k!+3∑k⩾0kk!+∑k⩾07k!−1=∑k⩾1k(k−1)!+2∑k⩾11(k−1)!+∑k⩾01k!+3∑k⩾11(k−1)!+7∑k⩾01k!−1=∑k⩾0k+1k!+2∑k⩾01k!+∑k⩾01k!+10∑k⩾01k!−1=∑k⩾11(k−1)!+∑k⩾01k!+2∑k⩾01k!+∑k⩾01k!+10∑k⩾01k!−1=15∑k⩾01k!−1=15e−1
Answered by Smail last updated on 22/Jan/20
∑∞n=1n4n!=∑∞n=1n3(n−1)(n−2)(n−3)(n−4)!n3(n−1)!=1(n−1)!+7(n−2)!+6(n−3)!+1(n−4)!S=∑∞n=4(1(n−1)!+7(n−2)!+6(n−3)!+1(n−4)!)+(1+162+272)S=452+∑∞n=41(n−1)!+∑∞n=47(n−2)!+∑∞n=46(n−3)!+∑∞n=41(n−4)!S=452+∑∞n=31n!+∑∞n=27n!+∑∞n=16n!+∑∞n=01n!=452+(e−1−1−12)+7(e−1−1)+6(e−1)+e=452+15e−52−14−6∑∞n=1n4n!=15e
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