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 ))](https://www.tinkutara.com/question/Q61111.png)
$$\mathrm{Please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{the}\:\mathrm{C}\:\mathrm{mean}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:…\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{mean}\:\mathrm{square}\:?? \\ $$$$\:\:\:\:\mathrm{I}\:\mathrm{mean}:\:\:\:\:\:\:\left(\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:…\:+\:\mathrm{n}\:\left(\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{also} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{1}} \right)^{\mathrm{2}} \:+\:\mathrm{2}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{3}\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:…\:+\:\mathrm{n}\:\left(\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{n}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what}\:\mathrm{the}\:\mathrm{2}\:\mathrm{on}\:\mathrm{C}\:\mathrm{represent}\:.\:\:\mathrm{Thanks}. \\ $$$$\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{2}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{3}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:…\:+\:\mathrm{n}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{2}} \:\:\:\:=\:\:\:\frac{\left(\mathrm{2n}\:−\:\mathrm{1}\right)!}{\left[\left(\mathrm{n}\:−\:\mathrm{1}\right)!\right]^{\mathrm{2}} } \\ $$
Commented by perlman last updated on 29/May/19
$${C}_{{n}} ^{{k}\:} =\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$$${withe}\:{n}!={n}\left({n}−\mathrm{1}\right)….\mathrm{2}×\mathrm{1} \\ $$$${mor}\:{generaly}\:\:\:\:\:{x}!=\Gamma\left({x}+\mathrm{1}\right)\:{gamma}\:{function}\:{defind}\:{for}\:{all}\:{z}\in{C}\:{withe}/{Re}\left({z}\right)>\mathrm{0} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:{and}\:\:\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 29/May/19
$$\mathrm{But}\:\mathrm{sir},\:\:\mathrm{how}\:\mathrm{come}\:\mathrm{we}\:\mathrm{have}\:\:\:\:\mathrm{3}.\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{2}} \:\:\mathrm{which}\:\mathrm{is}\:\:\:\:\mathrm{3}.\overset{\mathrm{2}} {\:}\mathrm{C}_{\mathrm{3}} \:\:\:\mathrm{is}\:\mathrm{this}\:\mathrm{correct} \\ $$
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](https://www.tinkutara.com/question/Q61133.png)
$$\left(\mathrm{1}+{x}\right)^{{n}} ={c}_{\mathrm{0}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{c}_{{n}} {x}^{{n}} \\ $$$$\frac{{d}}{{dx}}\left[\left(\mathrm{1}+{x}\right)^{{n}} \right]={c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{2}} {x}+\mathrm{3}{c}_{\mathrm{3}} {x}^{\mathrm{2}} +…+{nc}_{{n}} {x}^{{n}−\mathrm{1}} \\ $$$${x}×{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} ={c}_{\mathrm{1}} {x}+\mathrm{2}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} {x}^{\mathrm{3}} +..+{nc}_{{n}} {x}^{{n}} \\ $$$${now}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}} ={c}_{\mathrm{0}} +\frac{{c}_{\mathrm{1}} }{{x}}+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+…+\frac{{c}_{{n}} }{{x}^{{n}} } \\ $$$${now} \\ $$$$\left({c}_{\mathrm{1}} {x}+\mathrm{2}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +..+{nc}_{{n}} {x}^{{n}−\mathrm{1}} \right)×\left({c}_{\mathrm{0}} +\frac{{c}_{\mathrm{1}} }{{x}}+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+..+\frac{{c}_{{n}} }{{x}^{{n}} }\right) \\ $$$${take}\:{the}\:{terms}\:{independent}\:{of}\:{x} \\ $$$${c}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} ^{\mathrm{2}} +…+{nc}_{{n}} ^{\mathrm{2}} \\ $$$${left}\:{hand}\:{side}\: \\ $$$${x}×{n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} ×\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}} \\ $$$$={x}×{n}\left(\mathrm{1}+{x}^{} \right)^{{n}−\mathrm{1}} ×\frac{\left(\mathrm{1}+{x}\right)^{{n}} }{{x}^{{n}} } \\ $$$$={n}×{x}^{\mathrm{1}−{n}} ×\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$$${let}\:\left({r}+\mathrm{1}\right){th}\:{term}\:{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \:{contains} \\ $$$${x}^{{n}−\mathrm{1}} \:\:\:{so}\:{when}\:{multiplied}\:{by}\:{x}^{\mathrm{1}−{n}} \:\:{result}\:{is} \\ $$$${independent}\:{of}\:{x} \\ $$$$\mathrm{2}{n}−\mathrm{1}_{{C}_{{r}} ×\left({x}\right)^{{r}} } \\ $$$${x}^{{r}} ={x}^{{n}−\mathrm{1}} \:\:\:{so}\:{r}={n}−\mathrm{1} \\ $$$${now} \\ $$$${n}×\mathrm{2}{n}−\mathrm{1}_{{C}_{{n}−\mathrm{1}} } ×{x}^{{n}−\mathrm{1}} ×{x}^{\mathrm{1}−{n}} \\ $$$${n}×\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!×{n}!} \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!\left({n}−\mathrm{1}\right)!} \\ $$$${so} \\ $$$${c}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{3}{c}_{\mathrm{3}} ^{\mathrm{2}} +…+{nc}_{{n}} ^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!×\left({n}−\mathrm{1}\right)!} \\ $$$$\boldsymbol{{hence}}\:\boldsymbol{{proved}} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 29/May/19
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{But}\:\mathrm{that}\:\:\mathrm{C}^{\mathrm{2}} ,\:\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\:\mathrm{or}\:\:\mathrm{another}\:\mathrm{way}\:\mathrm{to}\:\mathrm{write}\:\:\:\overset{\mathrm{2}} {\:}\mathrm{C} \\ $$
Commented by Tawa1 last updated on 29/May/19
$$\mathrm{Ohh},\:\mathrm{i}\:\mathrm{get}\:\mathrm{sir}.\:\:\mathrm{It}\:\mathrm{is}\:\mathrm{square}.\:\:\mathrm{C}_{\mathrm{1}} \:×\:\mathrm{C}_{\mathrm{1}} \:\:=\:\:\mathrm{C}_{\mathrm{1}} ^{\mathrm{2}} \\ $$
Commented by tanmay last updated on 29/May/19
$${yes}\:{c}_{\mathrm{1}} ×{c}_{\mathrm{1}} =\left({c}_{\mathrm{1}} \right)^{\mathrm{2}} \:\:{square}\:\:{i}\:{have}\:{avoided}\:\left(.\right) \\ $$$${just}\:{written}\:{as}\:{c}_{\mathrm{1}} ^{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 29/May/19
$$\mathrm{Ohh},\:\mathrm{great}.\:\:\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$