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Question: If (1 + x)<sup>n</sup> = C<sub>0</sub> + C<sub>1</sub>x + C<sub>2</sub>x<sup>2</sup> + … + C<sub>n</...

If (1 + x)n = C0 + C1x + C2x2 + … + Cnxn, then the value of ∑0≤r<∑s≤n(r+s)\sum_{0 \leq r <}^{}{\sum_{s \leq n}^{}{(r + s)}} (Cr + Cs) is -

A

n2 · 2n

B

n · 2n

C

n2 · 22n

D

None of these

Answer

n2 · 2n

Explanation

Solution

We have, ∑r=0n∑s=0n(r+s)(Cr+Cs)\sum_{r = 0}^{n}{\sum_{s = 0}^{n}{(r + s)(C_{r} + C_{s})}}

= ∑r=0n∑s=0n(rCr+rCs+sCr+sCs)\sum_{r = 0}^{n}{\sum_{s = 0}^{n}{(rC_{r} + rC_{s} + sC_{r} + sC_{s})}}

= ∑r=0n[∑s=0nrCr+r∑s=0nCs+Cr∑s=0ns+∑s=0nsCs]\sum_{r = 0}^{n}\left\lbrack \sum_{s = 0}^{n}{rC_{r} + r}\sum_{s = 0}^{n}{C_{s} +}C_{r}\sum_{s = 0}^{n}{s +}\sum_{s = 0}^{n}{sC_{s}} \right\rbrack

= ∑r=0n[(n+1)r.Cr+r2n+n(n+1)2Cr+n.2n−1]\sum_{r = 0}^{n}\left\lbrack (n + 1)r.C_{r} + r2^{n} + \frac{n(n + 1)}{2}C_{r} + n.2^{n - 1} \right\rbrack

= (n + 1) (n · 2n–1) + 2n n(n+1)2\frac{n(n + 1)}{2}+ n(n+1)2\frac{n(n + 1)}{2}2n + n 2n–1 (n + 1)

= n (n + 1)2n + n (n + 1) 2n

= 2n (n + 1)2n …(1)

Also, ∑r=0n∑s=0n(r+s)\sum_{r = 0}^{n}{\sum_{s = 0}^{n}{(r + s)}} (Cr + Cs)

= ∑r=0n4r\sum_{r = 0}^{n}{4r}Cr + 2∑0≤r<∑s≤n(r+s)\sum_{0 \leq r <}^{}{\sum_{s \leq n}^{}{(r + s)}} (Cr + Cs)

\ 2n(n + 1) 2n = 4n · 2n–1 + 2 ∑0≤r<∑s≤n(r+s)\sum_{0 \leq r <}^{}{\sum_{s \leq n}^{}{(r + s)}} (Cr + Cs)

Ž ∑0≤r<∑s≤n(r+s)\sum_{0 \leq r <}^{}{\sum_{s \leq n}^{}{(r + s)}} (Cr + Cs) = n2 · 2n.

Hence (1) is correct answer.