Accordintoquestion,weneedsformula:∑n=n(n+1)2∑n2=n(n+1)(2n+1)6∑n3=n(n+1)23Now,fromnumerator:Ingeneralterm..........⇒N:1.n2+2(n−1)2+3(n−2)2+...+n.12∑nr=1(r)(n−(r−1))2⇒∑nr=1r[n2+r2−2r+1−2n(r−1)]⇒∑nr=1r[n2+r2−2r+1−2nr+2n]⇒∑nr=1r[r2−2r(1+n)+n2+1+2n]⇒∑nr=1[r3−2r2(1+n)+(n+1)2r]⇒∑nr=1r3−2(1+n)∑nr=1r2+(n+1)2∑nr=1rNow,(n)2(n+1)24−2(n+1)2n(2n+1)6+(n+1)2+1n2=(n+1)2[n24−n(2n−1)3+n(n+1)2]=(n+1)2n2[14−13(2+1n)+12(1+1n)]=(n+1)2n2[14−23+13n)+12+12n]And,fromDenomenator:Ingeneralterm..........D:13+23+33+...+n3|weknowsumofcubeis′n′naturalno.∑n3=n2(n+1)24Now,from:........ND=limn→∞⇒n2(n+1)2[14−23+13n+12+12n]n2(n+1)24limn→∞⇒[14−23+13n+12+12n]14|here:nisinity⇒34−2314=112×41=13Ans.