Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(y=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2y=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2y-y=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow y=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<1\)
ta có : 2y=\(\frac{1}{2}+\frac{1}{^{2^2}}+...+\frac{1}{2^{99}}\)
=> 2y-y=\(\frac{1}{2}-\frac{1}{2^{100}}\)
y=0,5=>y<1
Bài 1:
\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
......????
đặt A=1+1/1.2+1/2.3+...+1/49.50
ta có:y=1/1^2+1/2^2+...+1/50^2<A=1+1/1.2+1/2.3+...+1/49.50
mà A=1+1/1.2+1/2.3+...+1/49.50
=1+1-1/2+1/2-1/3+...+1/49-1/50
=2-1/50<2
=>y<2 (đpcm)
đặt A=1+1/1.2+1/2.3+...+1/49.50
ta có:y=1/1^2+1/2^2+...+1/50^2<A=1+1/1.2+1/2.3+...+1/49.50
mà A=1+1/1.2+1/2.3+...+1/49.50
=1+1-1/2+1/2-1/3+...+1/49-1/50
=2-1/50<2
=>y<2 (đpcm)