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Question Number 23856 by Tinkutara last updated on 08/Nov/17

The value of (C_0 + C_1 )(C_1 + C_2 ).... (C_(n−1) + C_n ) is (1) (((n + 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (2) (((n − 1)^n )/(n!)) ∙ C_1 C_2 .....C_n (3) (((n)^n )/((n + 1)!)) ∙ C_1 C_2 .....C_n (4) (((n)^n )/(n!)) ∙ C_1 C_2 .....C_n

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\left({C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} \right)\left({C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \right).... \\ $$$$\left({C}_{{n}−\mathrm{1}} \:+\:{C}_{{n}} \right)\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left({n}\:+\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\left({n}\:−\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\left({n}\right)^{{n}} }{\left({n}\:+\:\mathrm{1}\right)!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\left({n}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} .....{C}_{{n}} \\ $$

Commented by prakash jain last updated on 08/Nov/17

C_k +C_(k+1) =((n!)/(k!(n−k)!))+((n!)/((k+1)!(n−k−1)!)) =((n!)/(k!(n−k−1)!))[((n−k+k+1)/((k+1)(n−k)))] =((n+1!)/((k+1)!(n−k)!))=(((n+1)(n!))/((k+1)[k!(n−k)!])) =(((n+1))/((k+1)))^n C_k

$${C}_{{k}} +{C}_{{k}+\mathrm{1}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}−\mathrm{1}\right)!} \\ $$$$=\frac{{n}!}{{k}!\left({n}−{k}−\mathrm{1}\right)!}\left[\frac{{n}−{k}+{k}+\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({n}−{k}\right)}\right] \\ $$$$=\frac{{n}+\mathrm{1}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!}=\frac{\left({n}+\mathrm{1}\right)\left({n}!\right)}{\left({k}+\mathrm{1}\right)\left[{k}!\left({n}−{k}\right)!\right]} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)}\:^{{n}} {C}_{{k}} \\ $$

Commented by prakash jain last updated on 08/Nov/17

Π_(k=0) ^(n−1) (C_k +C_(k+1) ) =Π_(k=0) ^(n−1) ((n+1)/(k+1))C_k =(((n+1)^n )/(n!))C_0 C_1 ...C_(n−1) =(((n+1)^n )/(n!))C_1 C_2 ...C_n

$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({C}_{{k}} +{C}_{{k}+\mathrm{1}} \right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{n}+\mathrm{1}}{{k}+\mathrm{1}}{C}_{{k}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}{C}_{\mathrm{0}} {C}_{\mathrm{1}} ...{C}_{{n}−\mathrm{1}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}{C}_{\mathrm{1}} {C}_{\mathrm{2}} ...{C}_{{n}} \\ $$

Commented by prakash jain last updated on 08/Nov/17

Option (1) is correct.

$$\mathrm{Option}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{correct}. \\ $$

Commented by Tinkutara last updated on 09/Nov/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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