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A = 3/1×5 + 3/5×9 + 1/9×13 + ... + 9/97×101 + 3/101×105
A = 3/4 × (4/1×5 + 4/5×9 + 4/9×13 + ... + 4/97×101 + 4/101×105)
A = 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101 + 1/101 - 1/105)
A = 3/4 × (1 - 1/105)
A = 3/4 × 104/105
A = 26/35
B = 1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625
5B = 1 + 1/5 + 1/25 + 1/125 + 1/625 + 1/3125
5B - B = (1 + 1/5 + 1/25 + 1/125 + 1/625 + 1/3125) - (1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625)
4B = 1 - 1/15625
4B = 15624/15625
B = 15624/15625 : 4
B = 3906/15625
C = 1 + 2 + 4 + 8 + 16 + ... + 2048 + 4096
2C = 2 + 4 + 8 + 16 + 32 + ... + 4096 + 8192
2C - C = (2 + 4 + 8 + 16 + 32 + ... + 4096 + 8192) - (1 + 2 + 4 + 8 + ... + 2048 + 4096)
B = 8192 - 1
B = 8191
\(B=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{125\times129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{125}-\dfrac{1}{129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{129}\right)=\dfrac{1}{4}\times\dfrac{128}{129}=\dfrac{32}{129}\)
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
A = 4/1 x 5 + 4/5 x 9 + 4/9 x 13 + .... + 4/91 x 95 + 4/95 x 99
A = 1 - 1/5 + 1/5 -1/9 + 1/9 - 1/13 + .... + 1/91 - 1/95 + 1/95 - 1/99
A = 1 - 1/99
A = 98/99
B = 1/6 + 1/12 + 1/20 + ... + 1/132
B = 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + ... + 1/11 x 12
B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/11 - 1/12
B = 1/2 - 1/12
B = 5/12
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
\(D=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{21\cdot25}\)
\(4D=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{21\cdot25}\)
\(4D=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{21}-\dfrac{1}{25}\)
\(4D=1-\dfrac{1}{25}=\dfrac{24}{25}\)
\(D=\dfrac{24}{25}\cdot\dfrac{1}{4}=\dfrac{4\cdot6}{25\cdot4}=\dfrac{6}{25}\)