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Question

If (1+x)n=C0+C1x+C2x2+…+Cnxn, then the value of ∑∑0≤r<s≤n(r+s)(Cr+Cs) is

A
n22n
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B
n2n
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C
n222n
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D
None of these
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Solution

The correct option is A n2⋅2nWe have, ∑∑0≤r,s≤n(r+s)(Cr+Cs) =n∑r=0n∑s=0(rCr+rCs+sCr+sCs) =n∑r=0[n∑s=0rCr+n∑s=0rCs+n∑s=0sCr+n∑s=0sCs] =n∑r=0[(n+1)rCr+r2n+n(n+1)2Cr+n⋅2n−1] =(n+1)(n⋅2n−1)+2nn(n+1)2+n(n+1)22n+n⋅2n−1(n+1) =n(n+1)2n+n(n+1)2n =2n(n+1)2n Also ∑∑0≤r,s≤n(r+s)(Cr+Cs) =n∑r=04rCr+2∑∑0≤r<s≤n(r+s)(Cr+Cs) ∴2n(n+1)2n=4n⋅2n−1+2∑∑0≤r<s≤n(r+s)(Cr+Cs) ⇒∑∑0≤r<s≤n(r+s)(Cr+Cs)=n2⋅2n

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