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\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot
Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
\(\Rightarrow5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow5H-H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)\)
\(\Rightarrow4H=\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+..+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{11}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{11}}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{4.5^{11}}\)
\(\Rightarrow4H=\frac{1}{4}-\frac{1}{4.5^{11}}-\frac{11}{5^{12}}\)
\(\Rightarrow H=\frac{1}{16}-\frac{1}{4^2.5^{11}}-\frac{11}{4.5^{12}}\)
Ta có : \(5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow4H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}+\frac{11}{5^{12}}\)
\(\Rightarrow20H=1+\frac{1}{5}+...+\frac{1}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow16H=20H-4H=1+\frac{10}{5^{11}}-\frac{11}{5^{12}}\Leftrightarrow H=\frac{1+\frac{10}{5^{11}}-\frac{11}{5^{12}}}{16}.\)