Q.

For n≥2, let Cr= rn and an=∑r=0n 1Cr

Moderate

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A-P,B-R,C-Q,D-S

A-R,B-S,C-R,D-Q

A-R,B-S,C-Q,D-R

A-R,B-R,C-Q,D-Q

answer is D.

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Detailed Solution

using Cr=Cn−1, we get ∑r=0n rCr=∑r=0n rCn−r=∑r=0n n−rCr⇒2∑r=0n rCr=nan⇒∑r=0n rCr=12nansimilarly, ∑r=0n n−rCr=12nanusing, rCr=nCr−1 we get∑r=0n 1 rCr=1n∑r=1n 1Cr−1=1n∑r=0n 1Cr=1nan−1similarly ,∑r=0n−1 1(n−r)Cr=1nan−1

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