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Question

CMR

$C^{0}_{n}$ +

$C^{2}_{n}$ +

$C^{4}_{n}$ +...=

$C^{1}_{n}$ +

$C^{3}_{n}$ +

$C^{5}_{n}$ +...=

$2^{n-1}$

score 1 · Answer 1

Ta co:

$(1+x)^{n}=C^{0}_{n}+C^{1}_{n}.x+C^{2}_{n}.x^{2}+...+C^{k}_{n}.x^{k}+C^{n}_{n}.x^{n}$

Cho x=-1 ta duoc:

$0=C^{0}_{n}-C^{1}_{n}+C^{2}_{n}-C^{3}_{n}+...+C^{2k}_{n}-C^{2k+1}_{n}+...+(-1)^{n}C^{n}_{n}$

$\Rightarrow$

$C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...=C^{1}_{n}+C^{3}_{n}+C^{5}_{n}+...$ (1)

Cho x=1 ta co:

$2^{n}=C^{0}_{n}+C^{1}_{n}+C^{2}_{n}+C^{3}_{n}+....C^{2k}_{n}+C^{2k+1}_{n}+...+C^{n-1}_{n}+C^{n}_{n}$

Ap dung (1) ta co :

$2^{n}=2(C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...)$

$\Rightarrow$

$\frac{2^{n}}{2}=C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...$ hay

$2^{n-1}=C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...$ (2)

Tu (1) va(2)

$\Rightarrow$ dpcm

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