In the usual notations prove that (C0+C1)(C1+C2).....(Cn−1+Cn) =(n+1)nn!C1C2C3.....Cn
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Solution
We have prove that (C0+C1)C1−(C1+C2)C2......Cn−1+CnCn=(n+1)nn! or (C0C1+1)(C1C2+1).....(Cn−1Cn+1)=(n+1)nn! Now nCrnCr−1=n−r+1r=(n+1)r−1 ∴CrCr−1+1=(n+1)r or Cr+Cr−1Cr−1=(n+1)r Putting r = 1, 2, 3 ,........n in both sides and multiply etc Cn=C0