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Question Number 61111 by Tawa1 last updated on 29/May/19

Please what does the 2 on the C mean.          C_1 ^2  + 2 C_2 ^2  + 3 C_3 ^2  + ... + n C_n ^2     =   (((2n − 1)!)/([(n − 1)!]^2 ))  Does the 2 on C mean square ??      I mean:      (C_1 )^2  + 2(C_2 )^2  + 3(C_3 )^2  + ... + n (C_n )^2   which is also              ( ^n C_1 )^2  + 2( ^n C_2 )^2  + 3( ^n C_3 )^2  + ... + n ( ^n C_n )^2     I just want to know what the 2 on C represent .  Thanks.        C_1 ^2  + 2 C_2 ^2  + 3 C_3 ^2  + ... + n C_n ^2     =   (((2n − 1)!)/([(n − 1)!]^2 ))

$$\mathrm{Please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{the}\:\mathrm{C}\:\mathrm{mean}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{mean}\:\mathrm{square}\:?? \\ $$$$\:\:\:\:\mathrm{I}\:\mathrm{mean}:\:\:\:\:\:\:\left(\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\left(\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{also} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{represent}\:.\:\:\mathrm{Thanks}. \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:...\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$

Commented by perlman last updated on 29/May/19

C_n ^(k ) =((n!)/(k!(n−k)!))  withe n!=n(n−1)....2×1  mor generaly     x!=Γ(x+1) gamma function defind for all z∈C withe/Re(z)>0  Γ(x+1)=xΓ(x) and  Γ(1)=1

$${C}_{{n}} ^{{k}\:} =\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$$${withe}\:{n}!={n}\left({n}−\mathrm{1}\right)....\mathrm{2}×\mathrm{1} \\ $$$${mor}\:{generaly}\:\:\:\:\:{x}!=\Gamma\left({x}+\mathrm{1}\right)\:{gamma}\:{function}\:{defind}\:{for}\:{all}\:{z}\in{C}\:{withe}/{Re}\left({z}\right)>\mathrm{0} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:{and}\:\:\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 29/May/19

But sir,  how come we have    3. C_3 ^2   which is    3. ^2 C_3    is this correct

$$\mathrm{But}\:\mathrm{sir},\:\:\mathrm{how}\:\mathrm{come}\:\mathrm{we}\:\mathrm{have}\:\:\:\:\mathrm{3}.\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:\:\mathrm{which}\:\mathrm{is}\:\:\:\:\mathrm{3}.\overset{\mathrm{2}} {\:}\mathrm{C}_{\mathrm{3}} \:\:\:\mathrm{is}\:\mathrm{this}\:\mathrm{correct} \\ $$

Answered by tanmay last updated on 29/May/19

(1+x)^n =c_0 +c_1 x+c_2 x^2 +...+c_n x^n   (d/dx)[(1+x)^n ]=c_1 +2c_2 x+3c_3 x^2 +...+nc_n x^(n−1)   x×n(1+x)^(n−1) =c_1 x+2c_2 x^2 +3c_3 x^3 +..+nc_n x^n   now (1+(1/x))^n =c_0 +(c_1 /x)+(c_2 /x^2 )+...+(c_n /x^n )  now  (c_1 x+2c_2 x^2 +..+nc_n x^(n−1) )×(c_0 +(c_1 /x)+(c_2 /x^2 )+..+(c_n /x^n ))  take the terms independent of x  c_1 ^2 +2c_2 ^2 +3c_3 ^2 +...+nc_n ^2   left hand side   x×n(1+x)^(n−1) ×(1+(1/x))^n   =x×n(1+x^ )^(n−1) ×(((1+x)^n )/x^n )  =n×x^(1−n) ×(1+x)^(2n−1)   let (r+1)th term of (1+x)^(2n−1)  contains  x^(n−1)    so when multiplied by x^(1−n)   result is  independent of x  2n−1_(C_r ×(x)^r )   x^r =x^(n−1)    so r=n−1  now  n×2n−1_C_(n−1)  ×x^(n−1) ×x^(1−n)   n×(((2n−1)!)/((n−1)!×n!))  =(((2n−1)!)/((n−1)!(n−1)!))  so  c_1 ^2 +2c_2 ^2 +3c_3 ^2 +...+nc_n ^2 =(((2n−1)!)/((n−1)!×(n−1)!))  hence proved

$$\left(\mathrm{1}+{x}\right)^{{n}} ={c}_{\mathrm{0}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} {x}^{\mathrm{2}} +...+{c}_{{n}} {x}^{{n}} \\ $$$$\frac{{d}}{{dx}}\left[\left(\mathrm{1}+{x}\right)^{{n}} \right]={c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{2}} {x}+\mathrm{3}{c}_{\mathrm{3}} {x}^{\mathrm{2}} +...+{nc}_{{n}} {x}^{{n}−\mathrm{1}} \\ $$$${x}×{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} ={c}_{\mathrm{1}} {x}+\mathrm{2}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} {x}^{\mathrm{3}} +..+{nc}_{{n}} {x}^{{n}} \\ $$$${now}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}} ={c}_{\mathrm{0}} +\frac{{c}_{\mathrm{1}} }{{x}}+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+...+\frac{{c}_{{n}} }{{x}^{{n}} } \\ $$$${now} \\ $$$$\left({c}_{\mathrm{1}} {x}+\mathrm{2}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +..+{nc}_{{n}} {x}^{{n}−\mathrm{1}} \right)×\left({c}_{\mathrm{0}} +\frac{{c}_{\mathrm{1}} }{{x}}+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+..+\frac{{c}_{{n}} }{{x}^{{n}} }\right) \\ $$$${take}\:{the}\:{terms}\:{independent}\:{of}\:{x} \\ $$$${c}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} ^{\mathrm{2}} +...+{nc}_{{n}} ^{\mathrm{2}} \\ $$$${left}\:{hand}\:{side}\: \\ $$$${x}×{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} ×\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}} \\ $$$$={x}×{n}\left(\mathrm{1}+{x}^{} \right)^{{n}−\mathrm{1}} ×\frac{\left(\mathrm{1}+{x}\right)^{{n}} }{{x}^{{n}} } \\ $$$$={n}×{x}^{\mathrm{1}−{n}} ×\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \:{contains} \\ $$$${x}^{{n}−\mathrm{1}} \:\:\:{so}\:{when}\:{multiplied}\:{by}\:{x}^{\mathrm{1}−{n}} \:\:{result}\:{is} \\ $$$${independent}\:{of}\:{x} \\ $$$$\mathrm{2}{n}−\mathrm{1}_{{C}_{{r}} ×\left({x}\right)^{{r}} } \\ $$$${x}^{{r}} ={x}^{{n}−\mathrm{1}} \:\:\:{so}\:{r}={n}−\mathrm{1} \\ $$$${now} \\ $$$${n}×\mathrm{2}{n}−\mathrm{1}_{{C}_{{n}−\mathrm{1}} } ×{x}^{{n}−\mathrm{1}} ×{x}^{\mathrm{1}−{n}} \\ $$$${n}×\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!×{n}!} \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!\left({n}−\mathrm{1}\right)!} \\ $$$${so} \\ $$$${c}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} ^{\mathrm{2}} +...+{nc}_{{n}} ^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!×\left({n}−\mathrm{1}\right)!} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{proved}} \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 29/May/19

Wow, God bless you sir.  But that  C^2 ,   is the 2 square  or  another way to write    ^2 C

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{But}\:\mathrm{that}\:\:\mathrm{C}^{\mathrm{2}} ,\:\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\:\mathrm{or}\:\:\mathrm{another}\:\mathrm{way}\:\mathrm{to}\:\mathrm{write}\:\:\:\overset{\mathrm{2}} {\:}\mathrm{C} \\ $$

Commented by Tawa1 last updated on 29/May/19

Ohh, i get sir.  It is square.  C_1  × C_1   =  C_1 ^2

$$\mathrm{Ohh},\:\mathrm{i}\:\mathrm{get}\:\mathrm{sir}.\:\:\mathrm{It}\:\mathrm{is}\:\mathrm{square}.\:\:\mathrm{C}_{\mathrm{1}} \:×\:\mathrm{C}_{\mathrm{1}} \:\:=\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \\ $$

Commented by tanmay last updated on 29/May/19

yes c_1 ×c_1 =(c_1 )^2   square  i have avoided (.)  just written as c_1 ^2

$${yes}\:{c}_{\mathrm{1}} ×{c}_{\mathrm{1}} =\left({c}_{\mathrm{1}} \right)^{\mathrm{2}} \:\:{square}\:\:{i}\:{have}\:{avoided}\:\left(.\right) \\ $$$${just}\:{written}\:{as}\:{c}_{\mathrm{1}} ^{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 29/May/19

Ohh, great.  Thanks sir. I appreciate

$$\mathrm{Ohh},\:\mathrm{great}.\:\:\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

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