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Question: If C<sub>r</sub> = <sup>n</sup>C<sub>r</sub> and (C<sub>0</sub> + C<sub>1</sub>) (C<sub>1</sub> + C<...

If Cr = nCr and (C0 + C1) (C1 + C2) ………

(Cn–1 + Cn) = k (n+1)nn!\frac { ( \mathrm { n } + 1 ) ^ { \mathrm { n } } } { \mathrm { n } ! } , then the value of k is

A

C0C1C2 ….. Cn

B
C

C1 + C2 + C3 + ….+ Cn

D

None of these

Answer

C0C1C2 ….. Cn

Explanation

Solution

C1.C2.C3…..Cn (1+C0C1)(1+C1C2)\left( 1 + \frac { \mathrm { C } _ { 0 } } { \mathrm { C } _ { 1 } } \right) \left( 1 + \frac { \mathrm { C } _ { 1 } } { \mathrm { C } _ { 2 } } \right) …..(1+Cn−1Cn)\left( 1 + \frac { C _ { n - 1 } } { C _ { n } } \right)

= C1.C2.C3….Cn (n+1n)(n+1n−1)\left( \frac { \mathrm { n } + 1 } { \mathrm { n } } \right) \left( \frac { \mathrm { n } + 1 } { \mathrm { n } - 1 } \right) …..

= C1.C2.C3…...Cn

Ž k = C0.C1.C2…Cn Q C0 = 1