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