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Question

If n+1C2+2(2C2+3C2+4C2+.....+nC2)=12+22+32+....+1002 then find sum of digits of n ?

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Solution

L.H.S.
=n+1C2+2(2C2+3C2+4C2+.......+nC2)
=n+1C2+2(3C3+3C2+4C2+.......+nC2)
=n+1C2+2.n+1C3(nCr+nCr1=n+1Cr)
=n+1C2+n+1C3+n+1C3
=n+2C3+n+1C3
=>(n+2)!3!(n1)!+(n+1)!3!(n2)!=n(n+1)(2n+1)6
[ L.H.S. = R.H.S ] .
n(n+1)(2n+1)6=100(101)(201)6n=100

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