CM:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)<1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
https://olm.vn/thanhvien/2623944 xem câu trả lời của câu hỏi của nick này
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}< 1\)
hay \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)
\(=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)
\(=\frac{1}{2}+\frac{1}{101}\)
mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)
=> đ p c m
1/2^2=1/2.2<1/1.2
1/3^2=1/3.3<1/2.3
....
1/100^2=1/100.100<1/99.100
do đó 1/2^2+1/3^2+....+1/100^2<1/1.2+1/2.3+...+1/99.100
Áp dụng công thức :1/n - 1/n+1=(n+1-n)/n.(n+1)=1/n(n+1)
=>1/2^2+1/3^2+...+1/100^2<1/1-1/2+1/2-1/3+...+1/99-1/100=1/1-1/100=99/100<1
=>1/2^2+1/3^2+....+1/100^2<1(đpcm)
Ta có : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
.........
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....\frac{1}{100^2}<1-\frac{1}{100}=\frac{99}{100}\)
\(Mà\frac{99}{100}<1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)