[toán 11] bài tập về nhị thức niu-tơn

N

nguyenbahiep1

[TEX]{C}_{2n+1}^{0} + {C}_{2n+1}^{2} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n} = {2}^{n}[/TEX]


ta tính

[laTEX] (1+1)^{2n+1} = 2^{2n+1} = C_{2n+1}^0 + C_{2n+1}^1 + ...+ C_{2n+1}^{2n}+ C_{2n+1}^{2n+1} \\ \\ A = C_{2n+1}^0 + ..+ C_{2n+1}^{2n} \\ \\ B = C_{2n+1}^1 + C_{2n+1}^{2n+1} \\ \\ A + B = 2^{2n+1} \\ \\ ( 1-1)^{2n+1} = A - B = 0 \Rightarrow A = B \\ \\ 2A = 2^{2n+1} \Rightarrow A = 2^{2n} \Rightarrow dpcm[/laTEX]
 
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T

thien0526

Đề sai thì phải: Vế phải phải bằng [TEX]2^{2n}[/TEX] chứ???

Đặt [TEX]A= C_{2n+1}^0+ C_{2n+1}^2+ C_{2n+1}^4+...+C_{2n+1}^{2n-2} +C_{2n+1}^{2n}[/TEX] (1)

Ta cũng có [TEX]A= C_{2n+1}^{2n+1}+ C_{2n+1}^{2n-1}+ C_{2n+1}^{2n-3}+..+ C_{2n+1}^3+C_{2n+1}^1[/TEX] (2)

Cộng (1) và (2) vế theo vế, ta được:

[TEX]2A= C_{2n+1}^0+ C_{2n+1}^1+ C_{2n+1}^2+ C_{2n+1}^3+C_{2n+1}^4+...+ C_{2n+1}^{2n-1}+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{2n+1}[/TEX]

\Rightarrow[TEX]A=\frac{2^{2n+1}}{2}=2^{2n}[/TEX](đpcm)
 
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M

mavuongkhongnha

Chứng minh:
[TEX]{C}_{2n+1}^{0} + {C}_{2n+1}^{2} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n} = {2}^{n}[/TEX]

ta có :

[TEX](1+1)^{2n+1}={C}_{2n+1}^{0} +{C}_{2n+1}^{1} + {C}_{2n+1}^{2} +{C}_{2n+1}^{3} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n}+{C}_{2n+1}^{2n+1} [/tex]

[TEX]{C}_{2n+1}^{0} -{C}_{2n+1}^{1} + {C}_{2n+1}^{2} -{C}_{2n+1}^{3} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n}-{C}_{2n+1}^{2n+1} [/TEX]

[TEX]=> (1+1)^{2n+1}+(1-1)^{2n+1}=2[{C}_{2n+1}^{0} + {C}_{2n+1}^{2} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n}] [/TEX]

[TEX]{C}_{2n+1}^{0} + {C}_{2n+1}^{2} + {C}_{2n+1}^{4} +...+ {C}_{2n+1}^{2n}=2^{2n}[/TEX]

sao vậy nhỉ hình như tớ sai ở đâu ý :-SS
 
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