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A=1/1^2+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2
A=1+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2
A<1+(1/2^2+1/2.3+1/3/4+...+1/98.99+1/99.100) (giữ nguyên phân số 1/2^2)
A<1+ (1/4+1/2-1/3+1/3-1/4+...+1/99-1/99+1/99-1/100)
A<1+ (1/4+1/2-1/100)
Mà 1/4+1/2-1/100 <1/4+1/2=3/4
=>A<1+3/4=7/4
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Sửa đề: A=1/2^2+...+1/100^2
1/2^2<1/1*2
1/3^2<1/2*3
...
1/100^2<1/99*100
=>A<1-1/2+1/2-1/3+...+1/99-1/100
=>A<99/100<1
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Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1\)
Vậy.......
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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 \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
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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 : \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
.........
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....\frac{1}{100^2}<1-\frac{1}{100}=\frac{99}{100}\)
\(Mà\frac{99}{100}<1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)
Vì \(\dfrac{1}{a}\left(a>1\right)< 1với\forall a\)
mà \(2^2;3^2;.....;100^2>1\)
\(=>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)