Question Number 23856 by Tinkutara last updated on 08/Nov/17
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\left({C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} \right)\left({C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \right)…. \\ $$$$\left({C}_{{n}−\mathrm{1}} \:+\:{C}_{{n}} \right)\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left({n}\:+\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} …..{C}_{{n}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\left({n}\:−\:\mathrm{1}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} …..{C}_{{n}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\left({n}\right)^{{n}} }{\left({n}\:+\:\mathrm{1}\right)!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} …..{C}_{{n}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\left({n}\right)^{{n}} }{{n}!}\:\centerdot\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} …..{C}_{{n}} \\ $$
Commented by prakash jain last updated on 08/Nov/17
$${C}_{{k}} +{C}_{{k}+\mathrm{1}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}−\mathrm{1}\right)!} \\ $$$$=\frac{{n}!}{{k}!\left({n}−{k}−\mathrm{1}\right)!}\left[\frac{{n}−{k}+{k}+\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({n}−{k}\right)}\right] \\ $$$$=\frac{{n}+\mathrm{1}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!}=\frac{\left({n}+\mathrm{1}\right)\left({n}!\right)}{\left({k}+\mathrm{1}\right)\left[{k}!\left({n}−{k}\right)!\right]} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)}\:^{{n}} {C}_{{k}} \\ $$
Commented by prakash jain last updated on 08/Nov/17
$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({C}_{{k}} +{C}_{{k}+\mathrm{1}} \right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{{n}+\mathrm{1}}{{k}+\mathrm{1}}{C}_{{k}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}{C}_{\mathrm{0}} {C}_{\mathrm{1}} …{C}_{{n}−\mathrm{1}} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}{C}_{\mathrm{1}} {C}_{\mathrm{2}} …{C}_{{n}} \\ $$
Commented by prakash jain last updated on 08/Nov/17
$$\mathrm{Option}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by Tinkutara last updated on 09/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$