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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

$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \left({x}+\mathrm{1}\right)^{{n}} \\ $$$$=\left(^{{n}} {C}_{\mathrm{0}} {x}^{\mathrm{0}} +\:^{{n}} {C}_{\mathrm{1}} {x}+...+\:^{{n}} {C}_{{n}} {x}^{{n}} \right)Ɨ \\ $$$$\:\:\:\:\:\left(\:^{{n}} {C}_{\mathrm{0}} {x}^{{n}} +\:^{{n}} {C}_{\mathrm{1}} {x}^{{n}āˆ’\mathrm{1}} +...+^{{n}} {C}_{{n}} {x}^{\mathrm{0}} \right) \\ $$$${coeffecient}\:{of}\:{x}^{{n}} {in}\:{lhs}=\:^{\mathrm{2}{n}} {C}_{{n}} \\ $$$${coeffecient}\:{of}\:{x}^{{n}} {in}\:{rhs}= \\ $$$$\:\:\:{C}_{\mathrm{0}} ^{\mathrm{2}} +{C}_{\mathrm{1}} ^{\mathrm{2}} +..+{C}_{{n}} ^{\mathrm{2}} \\ $$$${hence} \\ $$$$\:\:\:{C}_{\mathrm{0}} ^{\mathrm{2}} +{C}_{\mathrm{1}} ^{\mathrm{2}} +..+{C}_{{n}} ^{\mathrm{2}} =^{\mathrm{2}{n}} {C}_{{n}} \\ $$$$ \\ $$

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