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Question Number 54740 by gunawan last updated on 10/Feb/19

Prove that 1. ((n),(r) ) = (((n−1)),(( r)) ) + (((n−1)),(( r−1)) ) 2. ((n),(r) ) + ((( n)),((r−1)) ) = (((n+1)),(( r)) ) 3. ((n),(0) )+ ((n),(1) )+ ((n),(2) )+..+ ((n),(n) )=2^n

$${Prove}\:{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:{r}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\: \\ $$$$\mathrm{3}.\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+..+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} \\ $$

Answered by Kunal12588 last updated on 10/Feb/19

3.^n C_0 +^n C_1 +^n C_2 +...+^n C_n =2^n eq^n 1 from binomial expansion (a+b)^p =^p C_0 a^p +^p C_1 a^(p−1) b^1 +...+^p C_r a^(p−r) b^r +...+^p C_p b^p [r<p] comparing with the question ∵ 1^(anything) =1 eq^n 1 can be written as : ^n C_0 1^n +^n C_1 1^(n−1) 1^1 +^n C_2 1^(n−2) 1^2 +...+^n C_n 1^n =(1+1)^n =2^n proved

$$\mathrm{3}.\:^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{1}} +^{{n}} {C}_{\mathrm{2}} +...+^{{n}} {C}_{{n}} =\mathrm{2}^{{n}} \:\:\:\:\:\:\:\:{eq}^{{n}} \:\mathrm{1} \\ $$$${from}\:{binomial}\:{expansion} \\ $$$$\left({a}+{b}\right)^{{p}} =^{{p}} {C}_{\mathrm{0}} {a}^{{p}} +^{{p}} {C}_{\mathrm{1}} {a}^{{p}−\mathrm{1}} {b}^{\mathrm{1}} +...+^{{p}} {C}_{{r}} {a}^{{p}−{r}} {b}^{{r}} +...+^{{p}} {C}_{{p}} {b}^{{p}} \:\:\:\:\:\:\:\:\left[{r}<{p}\right] \\ $$$${comparing}\:{with}\:{the}\:{question} \\ $$$$\because\:\mathrm{1}^{{anything}} =\mathrm{1} \\ $$$${eq}^{{n}} \:\:\mathrm{1}\:{can}\:{be}\:{written}\:{as}\::\: \\ $$$$\:^{{n}} {C}_{\mathrm{0}} \mathrm{1}^{{n}} +^{{n}} {C}_{\mathrm{1}} \mathrm{1}^{{n}−\mathrm{1}} \mathrm{1}^{\mathrm{1}} +^{{n}} {C}_{\mathrm{2}} \mathrm{1}^{{n}−\mathrm{2}} \mathrm{1}^{\mathrm{2}} +...+^{{n}} {C}_{{n}} \mathrm{1}^{{n}} \\ $$$$=\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\mathrm{2}^{{n}} \:\:\:\:\:{proved} \\ $$

Commented by gunawan last updated on 10/Feb/19

nice solution thank you Sir

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

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

another way (1+x)^n =nc_0 +nc_1 x+...+nc_n x^n put x=1 2^n =nc_0 +nc_1 +...+nc_n

$${another}\:{way}\:\left(\mathrm{1}+{x}\right)^{{n}} ={nc}_{\mathrm{0}} +{nc}_{\mathrm{1}} {x}+...+{nc}_{{n}} {x}^{{n}} \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\mathrm{2}^{{n}} ={nc}_{\mathrm{0}} +{nc}_{\mathrm{1}} +...+{nc}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 11/Feb/19

3) let p(x) =Σ_(k=0) ^n C_n ^k x^k =(x+1)^n x=1 ⇒ Σ_(k=0) ^n C_n ^k =2^n ⇒C_n ^0 +C_n ^1 +C_n ^2 +....+C_n ^n =2^n .

$$\left.\mathrm{3}\right)\:{let}\:{p}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:=\left({x}+\mathrm{1}\right)^{{n}} \:\:\: \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:=\mathrm{2}^{{n}} \:\Rightarrow{C}_{{n}} ^{\mathrm{0}} \:+{C}_{{n}} ^{\mathrm{1}} \:+{C}_{{n}} ^{\mathrm{2}} \:+....+{C}_{{n}} ^{{n}} \:\:=\mathrm{2}^{{n}} \:. \\ $$

Answered by Kunal12588 last updated on 10/Feb/19

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

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

Answered by Kunal12588 last updated on 10/Feb/19

2.^n C_r +^n C_(r−1) =^(n+1) C_r LHS =((n!)/(r!(n−r)!)) + ((n!)/((r−1)!(n−r+1)!)) =((n!)/((r−1)!(n−r)!))((1/r)+(1/(n−r+1))) =(((n+1)!)/(r!(n+1−r)!))=^(n+1) C_r =RHS proved

$$\mathrm{2}.\:^{{n}} {C}_{{r}} +^{{n}} {C}_{{r}−\mathrm{1}} =^{{n}+\mathrm{1}} {C}_{{r}} \\ $$$${LHS} \\ $$$$=\frac{{n}!}{{r}!\left({n}−{r}\right)!}\:+\:\frac{{n}!}{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!} \\ $$$$=\frac{{n}!}{\left({r}−\mathrm{1}\right)!\left({n}−{r}\right)!}\left(\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{n}−{r}+\mathrm{1}}\right) \\ $$$$=\frac{\left({n}+\mathrm{1}\right)!}{{r}!\left({n}+\mathrm{1}−{r}\right)!}=^{{n}+\mathrm{1}} {C}_{{r}} ={RHS}\:\:{proved} \\ $$

Commented by Kunal12588 last updated on 10/Feb/19

same as (1) just replace (n−1) with n

$${same}\:{as}\:\left(\mathrm{1}\right)\:{just}\:{replace}\:\left({n}−\mathrm{1}\right)\:{with}\:{n} \\ $$

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