S= (1+1/4).(1+1/9).(1+1/16).(1+1/25). ... .(1+1/121).(1+1/144)
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\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\)( đpcm )
\(=1^2+2^2+3^2+4^2+...+12^2+13^2\\ =\dfrac{13\times\left(13+1\right)\times\left(13\times2+1\right)}{6}=\dfrac{13\times14\times27}{6}=819\)
có
1x1=1
2x2=4
3x3=9
......
121=11x11
144=12x12
169=13x13
các số còn thiếu là
4x4=16
5x5=25
6x6=36
7x7=49
8x8=64
9x9=81
10x10=100
có
1+4+9+16+25+36+49+64+81+100+121+144+169
=819
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)....\left(\frac{1}{121}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.\frac{-24}{25}....\frac{-120}{121}\)
\(=\left[\left(-1\right)\left(-1\right)\left(-1\right)\left(-1\right)....\left(-1\right)\left(10\right)\text{thừa số -1 }\right].\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{10.12}{11.11}\)
\(=\frac{1.12}{2.11}=\frac{6}{11}\)
Sửa đề : \(S=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{144}\right)\)
\(\Rightarrow S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{143}{144}\)
\(\Rightarrow S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{11.13}{12.12}\)
\(\Rightarrow S=\frac{1.2.3.....11}{2.3.4.....12}.\frac{3.4.5.....13}{2.3.4.....12}\)
\(\Rightarrow S=\frac{1}{12}.\frac{13}{2}\)
\(\Rightarrow S=\frac{13}{24}\)