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a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2S=2+1+...+\frac{1}{2^{99}}\)
\(2S-S=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(S=2-\frac{1}{2^{100}}\)
phần b tương tự
a. S=1+1/2+1/2^2+1/2^3+...+1/2^100
2S=2+1+1/2+1/2^2+...+1/2^99
2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)
S=2-1/2^100
S=2^101-1/2^100
a) Đặt M=1/2+1/22+1/23+...+1/21998
=>2M=1+1/2+1/22+1/23+...+1/21997
2M-M=(1+1/2+1/22+1/23+...+1/21997)-(1/2+1/22+1/23+...+1/21998)
M=1-1/21998
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\frac{4949}{9900}\)
\(A=\frac{4949}{19800}\)
\(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+....+\frac{3}{1+2+3+...+100}\)
\(=3+\frac{3}{2.3:2}+\frac{3}{3.4:2}+...+\frac{3}{100.101:2}\)
\(=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6\left(1-\frac{1}{101}\right)=\frac{6.100}{101}=\frac{600}{101}\)