[Toán 8]- Rút gọn P

T

trinhminh18

$P= \dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+ \dfrac{10}{2^{10}}$
\Rightarrow$\dfrac{1}{2}P=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{10}{2^{11}}$
\Rightarrow$P-\dfrac{1}{2}P=\dfrac{1}{2}P$=$\dfrac{1}{2}+\dfrac{1}{2^2}
+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}-\dfrac{10}{2^{11}}$
Đặt S=$\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}$+ $\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}$
\Rightarrow$2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}$+ $\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^9}$
\Rightarrow$2S-S=1-\dfrac{1}{2^{10}}$
Hay$S=\dfrac{2^{10}-1}{2^{10}}$
Lại có:
$\dfrac{1}{2}P=S-\dfrac{10}{2^{11}}=\dfrac{2^{10}-1}{2^{10}}-\dfrac{10}{2^{11}}=\dfrac{2^{11}-12}{2^{11}}$
 
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