Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c) gọi biểu thức là S = 2 + 2\(^2+2^3+.....+2^{50}\)
2S=2\(^2+2^3+2^4+......+2^{50}+2^{51}\)
\(2S-S=S=2^{51}-2\)
b) \(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{10}}\)
= \(2+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^9}\)
2S-S=S=(\(2+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^9}\))-( \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\))
bạn tự tìm S nhé
mink làm được như thế đó, phần a mink không muốn nhấn mỏi tay bạn ạ, đừng nghĩ mink ko biết làm nha
a/ Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...........+\dfrac{1}{n^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)
\(\Leftrightarrow A< 1\)
b/ Ta có :
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.................+\dfrac{1}{\left(2n\right)^2}\)
\(=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\right)\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{\left(n-1\right)n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
\(\Leftrightarrow B< \dfrac{1}{2}\)
\(\)\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(A< 1-\dfrac{1}{n}< 1\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2n^2}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B=\dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B< \dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{\left(n-1\right)n}\right)\)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
= \(\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+1-\dfrac{1}{4}+...+1-\dfrac{1}{100}}\)
= \(\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
=\(\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)= 2
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
Không phải chỉ là khi rút gọn mà trong khi thực hiện phép tính em cũng cần đưa về mẫu số dương em nhé.
Dạ em cảm ơn cô