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Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
gọi A là phân số thứ nhất, B là phân số thứ 2
\(\frac{A}{3}=\frac{3^{100}+1}{3^{100}+3}\)có phần bù là \(\frac{2}{3^{100}+3}\)
\(\frac{B}{3}=\frac{3^{99}+1}{3^{99}+3}\)có phần bù là \(\frac{2}{3^{99}+3}\)
ta thấy \(\frac{2}{3^{100}+3}< \frac{2}{3^{99}+3}\Rightarrow A>B\)
mink nghĩ vậy bạn ạ
\(A=\frac{3^{100}+1}{3^{99}+1}=\frac{\left(3^{99}+1\right)\times3-2}{3^{99}+1}=3-\frac{2}{3^{99}+1}\)
\(B=\frac{3^{99}+1}{3^{98}+1}=\frac{\left(3^{98}+1\right)\times3-2}{3^{98}+1}=3-\frac{2}{3^{98}+1}\)
Do 398 + 1 < 399 + 1
=> \(\frac{2}{3^{98}+1}>\frac{2}{3^{99}+1}\)
=> A > B