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1.3.5. ... .99=51/2.52/2. ... .100/2
nhân cả hai vế với 1.2...50.2^50, ta được
*vế 1
1.3.5. ... .99.1.2...50.2^50=1.3.5...99.2.2.2..2..1.2...50
=1.3.5...99.1.2.2.2.2.3.2.4.....2.50
1.3.....99.2.4..10=1.2.3.4.5...100 (1)
*vế 2
51/2.52/2. ... .100/2^50.1.2.3...50=51/2.52/2. ... .100/2.2.2...1.2.3...50
=(51/2).2.(52/2).2 ... .(100/2).2.....1.2.3...50
rút gọn ta sẽ đươc51.52.53...100.1.2.3...50(2)
từ (1) và (2)=>1.3.5. ... .99=51/2.52/2. ... .100/2
\(1.3.5....99=\frac{1.2.3.4....99.100}{2.4.6...100}=\frac{\left(1.2.3....50\right).\left(51.52.53...100\right)}{2^{50}.\left(1.2.3...50\right)}\)
\(=\frac{51.52.53....100}{2^{50}}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}......\frac{100}{2}\)
Ta có :
\(1.3.5.....99=\frac{1.2.3.4.....99.100}{2.4.6......100}\)
\(=\frac{1.2.3......99.100}{1.2.2.2.2.3......2.50}\)
\(=\frac{1.2.3.4......99.100}{2^{50}.1.2.3......50}\)
\(=\frac{51.52.....100}{2^{50}}\)
\(=\frac{51}{2}.\frac{52}{2}...........\frac{100}{2}\) (ĐPCM)
Đặt \(S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(S=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Ta có đpcm
