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Question Number 54646 by gunawan last updated on 08/Feb/19

Such That a. _(n+1) C_r =(((n+1). _n C_r )/((n−r+1))) b. _n C_0 +_n C_2 +_n C_(4...) =_n C_1 +_n C_3 +_n C_(5...) =2^(n−1)

$$\mathrm{Such}\:\mathrm{That} \\ $$$$\mathrm{a}.\:_{{n}+\mathrm{1}} {C}_{{r}} =\frac{\left({n}+\mathrm{1}\right).\:_{{n}} {C}_{{r}} }{\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\mathrm{b}.\:_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{4}...} =_{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{3}} +_{{n}} {C}_{\mathrm{5}...} =\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$ \\ $$

Answered by Kunal12588 last updated on 08/Feb/19

LHS=^(n+1) C_r =(((n+1)!)/(r!(n+1−r)!)) =(((n+1)n!)/(r!(n+1−r)(n−r)!))=((n+1)/(n−r+1))×((n!)/(r!(n−r)!)) =(((n+1) ^n C_r )/((n+1−r)))=RHS proved

$${LHS}=^{{n}+\mathrm{1}} {C}_{{r}} =\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!} \\ $$$$=\frac{\left({n}+\mathrm{1}\right){n}!}{{r}!\left({n}+\mathrm{1}−{r}\right)\left({n}−{r}\right)!}=\frac{{n}+\mathrm{1}}{{n}−{r}+\mathrm{1}}×\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\:\:^{{n}} {C}_{{r}} }{\left({n}+\mathrm{1}−{r}\right)}={RHS}\:\:\:{proved} \\ $$

Commented by gunawan last updated on 08/Feb/19

thank you Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Answered by Kunal12588 last updated on 10/Feb/19

b. from binimial expansion 0^n =(1−1)^n ⇒0=^n C_0 −^n C_1 +^n C_2 −.....+(−1)^n ∙^n C_n ⇒^n C_0 +^n C_2 +^n C_4 +....=^n C_1 +^n C_3 +^n C_5 +.... eq(1) LHS of eq(1) ^n C_0 +^n C_2 +^n C_4 +.... =^(n−1) C_0 +^(n−1) C_1 +^(n−1) C_2 +^(n−1) C_3 +^(n−1) C_4 +... =2^(n−1) proved I used the identities (1) and (3) from question 54740 see carefully

$${b}.\:\:{from}\:{binimial}\:{expansion} \\ $$$$\mathrm{0}^{{n}} =\left(\mathrm{1}−\mathrm{1}\right)^{{n}} \\ $$$$\Rightarrow\mathrm{0}=^{{n}} {C}_{\mathrm{0}} −^{{n}} {C}_{\mathrm{1}} +^{{n}} {C}_{\mathrm{2}} −.....+\left(−\mathrm{1}\right)^{{n}} \:\centerdot\:^{{n}} {C}_{{n}} \\ $$$$\Rightarrow^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{2}} +^{{n}} {C}_{\mathrm{4}} +....=^{{n}} {C}_{\mathrm{1}} +^{{n}} {C}_{\mathrm{3}} +^{{n}} {C}_{\mathrm{5}} +....\:\:\:\:{eq}\left(\mathrm{1}\right) \\ $$$${LHS}\:\:{of}\:{eq}\left(\mathrm{1}\right) \\ $$$$\:^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{2}} +^{{n}} {C}_{\mathrm{4}} +.... \\ $$$$=^{{n}−\mathrm{1}} {C}_{\mathrm{0}} +^{{n}−\mathrm{1}} {C}_{\mathrm{1}} +^{{n}−\mathrm{1}} {C}_{\mathrm{2}} +^{{n}−\mathrm{1}} {C}_{\mathrm{3}} +^{{n}−\mathrm{1}} {C}_{\mathrm{4}} +... \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} \:\:\:\:{proved} \\ $$$${I}\:{used}\:{the}\:{identities}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{3}\right)\:\:{from}\: \\ $$$${question}\:\mathrm{54740}\:{see}\:{carefully} \\ $$

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