Câu hỏi của Hoàng Huy Hoàn

Question

cho S =1-1/2 +1/3 -1/4 +....+1/2023-1/2024+1/2025 P=1/1013+1/1014+1/1015+.....+1/2024+1/2025 tính ( S - P ) ^2025

Akai Haruma · Accepted Answer

Lời giải:$S=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2025})-(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})\
=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2025})-2(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})$$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2025}-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1012}\
=\frac{1}{1013}+\frac{1}{1014}+...+\frac{1}{2025}\
=P$$\Rightarrow (S-P)^{2025}=0^{2025}=0$Câu trả lời: Lời giải:$S=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2025})-(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})\
=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2025})-2(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2024})$$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2025}-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1012}\
=\frac{1}{1013}+\frac{1}{1014}+...+\frac{1}{2025}\
=P$$\Rightarrow (S-P)^{2025}=0^{2025}=0$

Nguyễn Thanh Phong · Answer

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