Menu Close

If-1-x-n-C-0-C-1-x-C-2-x-2-C-n-x-n-then-C-1-2-C-2-2-C-n-2-is-equal-to-




Question Number 18762 by 786786AM last updated on 29/Jul/17
If (1+x)^n =C_0 +C_1 x+C_2 x^2 +...+C_n x^n ,  then C_1 ^2 +C_2 ^2 +....+C_n ^2  is equal to
$$\mathrm{If}\:\left(\mathrm{1}+{x}\right)^{{n}} ={C}_{\mathrm{0}} +{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{C}_{{n}} {x}^{{n}} , \\ $$$$\mathrm{then}\:{C}_{\mathrm{1}} \:^{\mathrm{2}} +{C}_{\mathrm{2}} \:^{\mathrm{2}} +….+{C}_{{n}} \:^{\mathrm{2}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Answered by Rishabh#1 last updated on 30/Jul/17
(1+x)^(2n) =(1+x)^n (x+1)^n   =(^n C_0 x^0 +^n C_1 x+...+^n C_n x^n )Ɨ       (^n C_0 x^n +^n C_1 x^(nāˆ’1) +...+^n C_n x^0 )  coeffecient of x^n in lhs=^(2n) C_n   coeffecient of x^n in rhs=     C_0 ^2 +C_1 ^2 +..+C_n ^2   hence     C_0 ^2 +C_1 ^2 +..+C_n ^2 =^(2n) C_n
$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \left({x}+\mathrm{1}\right)^{{n}} \\ $$$$=\left(^{{n}} {C}_{\mathrm{0}} {x}^{\mathrm{0}} +\:^{{n}} {C}_{\mathrm{1}} {x}+…+\:^{{n}} {C}_{{n}} {x}^{{n}} \right)Ɨ \\ $$$$\:\:\:\:\:\left(\:^{{n}} {C}_{\mathrm{0}} {x}^{{n}} +\:^{{n}} {C}_{\mathrm{1}} {x}^{{n}āˆ’\mathrm{1}} +…+^{{n}} {C}_{{n}} {x}^{\mathrm{0}} \right) \\ $$$${coeffecient}\:{of}\:{x}^{{n}} {in}\:{lhs}=\:^{\mathrm{2}{n}} {C}_{{n}} \\ $$$${coeffecient}\:{of}\:{x}^{{n}} {in}\:{rhs}= \\ $$$$\:\:\:{C}_{\mathrm{0}} ^{\mathrm{2}} +{C}_{\mathrm{1}} ^{\mathrm{2}} +..+{C}_{{n}} ^{\mathrm{2}} \\ $$$${hence} \\ $$$$\:\:\:{C}_{\mathrm{0}} ^{\mathrm{2}} +{C}_{\mathrm{1}} ^{\mathrm{2}} +..+{C}_{{n}} ^{\mathrm{2}} =^{\mathrm{2}{n}} {C}_{{n}} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *