Tính C = 1/1.2.3 + 1/2.3.4 +...+1/38.39.40
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A = 1 . 2 . 3 + 2 . 3 . 4 + ......... + 38 .39 . 40
\(\Rightarrow4A=1.2.3.4+2.3.4.4+......+38.39.40.4\)
\(\Rightarrow4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+.......+38.39.40.\left(41-37\right)\)
\(\Rightarrow4A=1.2.3.4+1.2.3.0+2.3.4.5-1.2.3.4+....+38.39.40.41-38.39.40.37\)
\(\Rightarrow4A=38.39.40.41\)
\(\Rightarrow A=\frac{38.39.40.41}{4}\)
\(\Rightarrow A=38.39.10.41\)
\(\Rightarrow A=607620\)
Vậy \(A=607620\)
A=1/2+1/4+1/8+1/16+1/32 A.2=1+1/2+1/4+1/8+1/16 A.2-A=A=1+1/2+1/4+1/8+1/16-1-1/2-1/4-1/6-1/8-1/16-1/32 A=1-1/32=31/32 B=1/2+1/2^2+1/2^3+...+1/2^2002 B.2=1+1/2+1/2^2+...+1/2^2001 B.2-B=B=1+1/2+...+1/2001-1/2-1/2^2-...-1/2^2002 B=1-1/2^2002=2^2002-1/2^2002 C=2/2.3.4+2/3.4.5+...+2/38.39.40 C=2.(1/2.3.4+1/3.4.5+...+1/38.39.40) C=2.1/2.(1/2.3 -1/3.4 +1/3.4 -1/4.5 +...+1/38.39 -1/39.40) C=1.(1/2.3 -1/39.40)=259/1560.
a. = 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8- 1/16 + 1/16 - 1/32 = 1-1/32=1/31
\(2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{98.99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{50.99-1}{100.99}=\dfrac{4949}{9900}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
= \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\)
= \(\frac{2-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{51-49}{49.50.51}\)
= \(\frac{1}{1.3}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
= \(\frac{1}{3}-\frac{1}{50.51}\)
= \(\frac{1}{3}-\frac{1}{2550}\)
= \(\frac{283}{850}\)
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{78.79.80}\)
\(\Rightarrow S=\dfrac{79\left(79+3\right)}{4\left(79+1\right)\left(79+2\right)}\)
\(S=\dfrac{79.82}{4.80.81}=\dfrac{79.41}{160.81}=\dfrac{3239}{12960}\)
nhân tổng trên cho 2 ta có;
2/1.2.3+2/2.3.4+.........+2/98.99.100
=1/1.2-1/2.3+1/2.3-1/3.4+........+1/98.99-1/99.100
=1/1.2-1/99.100
=4949/9900
/
Bạn ấn vào dòng chữ màu xanh Cho D = 1/1.2.3+1/2.3.4+....+1/37.38.39
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{38.39.40}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{38.39}-\frac{1}{39.40}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{39.40}\right)=\frac{1}{2}\cdot\frac{599}{1200}=\frac{599}{2400}\)