(C1 / C0) + 2 (C2 / C1) + 3 (C3 / C2) + …. + n (Cn / Cn-1) =
n(n-1)2
n(n+1)2
(n+1)(n+2)2
None of these
Explanation for correct option:
Step 1: Declaring the theorem
We know, nCr=n!(n-r)!r!
n!=n(n-1)!
Step 2: Solving the given equation
C1C0+2C2C1+3C3C2+....+nCnCn-1
Therefore, S = ∑r=1n(r)nCrnCr-1
⇒ S = ∑r=1n(r)n!(n-r)!r!n!(n-r+1)!(r-1)!
⇒S = ∑r=1n(r)(n-r+1)!(r-1)!(n-r)!r!
⇒ S = ∑r=1n(r)(n-r+1)!(n-r)!(r-1)!(n-r)!r(r-1)!
⇒ S = ∑r=1n(r)n-r+1r
⇒ S = ∑r=1n(n+1)-∑r=1nr
⇒ S = (n+1)∑(1)r=1n-∑r=1nr
⇒ S = (n+1)n-{1+2+3+4....n}
⇒ S = (n+1)n-n(n+1)2
⇒ S = n(n+1)2
Therefore, we can prove that C1C0+2C2C1+3C3C2+....+nCnCn-1 =n(n+1)2
Hence, Option(B) is correct.