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Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{3}{4}\left(đpcm\right)\)
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
3A-A=\(1-\frac{1}{3^{99}}\)
2A=\(1-\frac{1}{3^{99}}\)
vì 2A<1
=> A<\(\frac{1}{2}\)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
\(2A-A=1+\frac{2}{2}-\frac{1}{2}+\frac{3}{2^2}-\frac{2}{2^2}+...+\frac{100}{2^{99}}-\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Rightarrow A=2+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow A=2.\frac{1}{2^{100}}\)
Vậy \(A< 2\) do \(A=2\) nhân với một phân số nhỏ hơn \(1\)
\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)
\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)
\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)
\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)
\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)
\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)
\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)
\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)
Ta thấy đc quy luật:
\(\frac{2^2-1^2}{2^2}=\frac{2+1}{2+2}=\frac{3}{4}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}=\frac{6+2}{6+3}=\frac{8}{9}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}=\frac{12+3}{12+4}=\frac{15}{16}\)
Nên:
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}+...+\frac{100^2-99^2}{9900^2}=\frac{9900+99}{9900+100}=\frac{9999}{10000}\)
Hay A<1(đpcm)