Question

Cho \(S=1-\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\) và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)

Tính ( S - P )²⁰¹³

Answer 1

\(P=\frac{1}{1007}+\frac{1}{1008}+.....+\frac{1}{2012}+\frac{1}{2013}\)

\(P=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)

\(P=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)

\(\)
\(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2012}+\frac{1}{2013}=S\)

Vậy (S-P)²⁰¹³=0

Answer 2

(S-P)²⁰¹³=0