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Question Number 21655 by Joel577 last updated on 30/Sep/17

 (((2017)),((    0)) ) +  (((2017)),((    2)) ) +  (((2017)),((    4)) ) +  (((2017)),((    6)) ) + ... +  (((2017)),((2016)) )  is equal to ...

$$\begin{pmatrix}{\mathrm{2017}}\\{\:\:\:\:\mathrm{0}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{2017}}\\{\:\:\:\:\mathrm{2}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{2017}}\\{\:\:\:\:\mathrm{4}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{2017}}\\{\:\:\:\:\mathrm{6}}\end{pmatrix}\:+\:...\:+\:\begin{pmatrix}{\mathrm{2017}}\\{\mathrm{2016}}\end{pmatrix} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:... \\ $$

Answered by $@ty@m last updated on 30/Sep/17

(1+x)^n =^n C_0 +^n C_1 x+^n C_2 x^2 +...+^n C_n x^n   −−(1)  Put x=−1  0=^n C_0 −^n C_1 +^n C_2 +...+(−1)^n .^n C_n   ^n C_0 +^n C_2 +^n C_4 +...=^n C_1 +^n C_3 +^n C_5 +...   −−(2)  Put x=1 in (1)  (2)^n =^n C_0 +^n C_1 +^n C_2 +...+^n C_n   −−(3)  From (2) & (3)  ^n C_0 +^n C_2 +^n C_4 +...=^n C_1 +^n C_3 +^n C_5 +...  =2^(n−1)

$$\left(\mathrm{1}+{x}\right)^{{n}} =\:^{{n}} {C}_{\mathrm{0}} +\:^{{n}} {C}_{\mathrm{1}} {x}+\:^{{n}} {C}_{\mathrm{2}} {x}^{\mathrm{2}} +...+\:^{{n}} {C}_{{n}} {x}^{{n}} \:\:−−\left(\mathrm{1}\right) \\ $$$${Put}\:{x}=−\mathrm{1} \\ $$$$\mathrm{0}=\:^{{n}} {C}_{\mathrm{0}} −\:^{{n}} {C}_{\mathrm{1}} +\:^{{n}} {C}_{\mathrm{2}} +...+\left(−\mathrm{1}\right)^{{n}} .\:^{{n}} {C}_{{n}} \\ $$$$\:^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{2}} +\:^{{n}} {C}_{\mathrm{4}} +...=\:^{{n}} {C}_{\mathrm{1}} +\:^{{n}} {C}_{\mathrm{3}} +^{{n}} {C}_{\mathrm{5}} +...\:\:\:−−\left(\mathrm{2}\right) \\ $$$${Put}\:{x}=\mathrm{1}\:{in}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)^{{n}} =\:^{{n}} {C}_{\mathrm{0}} +\:^{{n}} {C}_{\mathrm{1}} +\:^{{n}} {C}_{\mathrm{2}} +...+\:^{{n}} {C}_{{n}} \:\:−−\left(\mathrm{3}\right) \\ $$$${From}\:\left(\mathrm{2}\right)\:\&\:\left(\mathrm{3}\right) \\ $$$$\:^{{n}} {C}_{\mathrm{0}} +^{{n}} {C}_{\mathrm{2}} +\:^{{n}} {C}_{\mathrm{4}} +...=\:^{{n}} {C}_{\mathrm{1}} +\:^{{n}} {C}_{\mathrm{3}} +^{{n}} {C}_{\mathrm{5}} +...\:\:=\mathrm{2}^{{n}−\mathrm{1}} \\ $$

Commented by Joel577 last updated on 30/Sep/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Commented by $@ty@m last updated on 30/Sep/17

Put n=2017  ^(2017) C_0 +^(2017) C_2 +^(2017) C_4 +...+^(2017) C_(2016) =2^(2016)  Ans.

$${Put}\:{n}=\mathrm{2017} \\ $$$$\:^{\mathrm{2017}} {C}_{\mathrm{0}} +\:^{\mathrm{2017}} {C}_{\mathrm{2}} +\:^{\mathrm{2017}} {C}_{\mathrm{4}} +...+\:^{\mathrm{2017}} {C}_{\mathrm{2016}} =\mathrm{2}^{\mathrm{2016}} \:{Ans}. \\ $$

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