1×22+2×32+3×42+⋯+n(n+1)212×2+22×3+32×4+⋯+n2(n+1)
Let S1=1×22+2×32+3×42+⋯+n(n+1)2
Tn=n(n+1)2 which is nth term of S1
So S1=∑(n3+2n2+n)
Let S2=12×2+22×3+32×4+⋯+n2(n+1)
Tn=n2(n+1) which is nth term of S2
So S2=∑(n3+n2)
∴1×22+2×32+3×42+⋯+n(n+1)212×2+22×3+32×4+⋯+n2(n+1)=∑(n3+2n2+n)∑(n3+n2)=14n2(n+1)2+26n(n+1)(2n+1)+12n(n+1)14n2(n+1)2+16n(n+1)(2n+1)=3n(n+1)+4(2n+1)+63n(n+1)+2(2n+1)=3n2+11n+103n2+7n+2=(3n+5)(n+2)(3n+1)(n+2)=3n+53n+1=1+43n+1
Now,
1+43n+1=87⇒n=9