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Question

If 1×22+2×32+3×42++n(n+1)212×2+22×3+32×4++n2(n+1)=87, then n=

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Solution

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=87n=9

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