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2S=2(1+2+22+...+250)
2S=2+22+...+251
2S-S=(2+22+...+251)-(1+2+22+...+250)
S=251-1<251
=>S<251
Ta có 1 - a2 = 1 - a + a - a2 = 1 - a + a(1 - a) = (1 - a)(1 + a)
Khi đó \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}...\frac{1-100^2}{100^2}\)
= \(\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}...\frac{\left(1-100\right)\left(1+100\right)}{100^2}\)
= \(-\frac{\left(2-1\right)\left(2+1\right).\left(3-1\right)\left(3+1\right)...\left(100-1\right)\left(100+1\right)}{2^2.3^2.4^2....100^2}\)
\(=-\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
\(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{31}.3^{61}}\)
\(=\frac{3^{61}\left(2^{50}+2^{90}\right)}{3^{61}\left(2^{51}+2^{31}\right)}\)
\(=\frac{2^{50}+2^{90}}{2^{51}+2^{31}}\)
\(=\frac{2^{31}\left(2^{19}+2^{59}\right)}{2^{31}\left(1+2^{20}\right)}\)
\(=\frac{2^{19}+2^{59}}{1+2^{20}}\)
\(S=1+2+2^2+...........+2^{50}\)
\(\Leftrightarrow2S=2+2^2+...........+2^{50}+2^{51}\)
\(\Leftrightarrow2S-S=\left(2+2^2+.........+2^{51}\right)-\left(1+2+2^2+..........+2^{50}\right)\)
\(\Leftrightarrow S=2^{51}-1\)
\(\Leftrightarrow S< 2^{51}\)
\(S=1+2+2^2+....+2^{50}\)
\(2S=2+2^2+2^3+....+2^{51}\)
\(2S-S=\left(2+2^2+2^3+...+2^{51}\right)-\left(1+2+2^2+...+2^{50}\right)\)
\(S=2^{51}-1\)
Vì \(2^{51}-1< 2^{51}\)
\(\Rightarrow S< 2^{51}\)
\(2S=2+2^2+.........+2^{51}\)
\(2S-S=\left(2+2^2+.......+2^{51}\right)-\left(1+2+.......+2^{50}\right)\)
\(\Rightarrow S=2^{51}-1< 2^{51}\)
Vậy S<251