Co the chung minh theo quy nap
+Voi $n=1$, ta co
$\dfrac{1}{1+1}=\dfrac{1}{{2^1}}$ (luon dung)
+Gia su (*) (de bai) dung voi $n=k$ (k thuoc N, k>1)
Suy ra $\dfrac{1.3.5...(2k-1)}{(k+1)(k+2)(k+3)...2k}=\dfrac{1}{{2^k}}$
Can chung minh (*) dung voi $n=k+1$
That vay:
$\dfrac{1.3.5...(2k-1)}{(k+1)(k+2)(k+3)...2k}=\dfrac{1}{{2^k}}$
\Leftrightarrow $\dfrac{1.3.5...(2k-1)}{(k+1)(k+2)(k+3)...2k}.\dfrac{k+1}{2k+2}=\dfrac{1}{{2^k}}.\dfrac{k+1}{2k+2}$
\Leftrightarrow $\dfrac{1.3.5...(2k-1)(2k+1)(k+1)}{(k+1)(k+2)(k+3)...2k(2k+1)(2k+2)}= \dfrac{1}{{2^k}}.\dfrac{1}{2}$
\Leftrightarrow $\dfrac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}= \dfrac{1}{{2^{k+1}}}$ (dpcm)
Suy ra dpcm
p/s:Khog co dau(
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