# Violympic toán 7

Cho S = $1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}$và P = $\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2012}+\dfrac{1}{2013}$. Tính$\left(P-S\right)^{2013}$

25 tháng 3 2017 lúc 21:15

Ta có: $S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}$

$=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)$

$=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)$

$=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)$

$=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}$

$\Rightarrow P-S=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\right)=0$

$\Rightarrow\left(P-S\right)^{2013}=0^{2013}=0$

Vậy $\left(P-S\right)^{2013}=0$

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