D = -1- 1/3- 1/6- 1/10- 1/15- .....- 1/1225 = ?
-1 - 1 phần 3 - 1 phần 6 - 1 phần 10 - 1 phần 15 - ... - 1 phần 1225
\(-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}=?\)
A = -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)
= -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))
Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)
Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))
= 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))
= 2×\(\dfrac{24}{50}\)
= \(\dfrac{24}{25}\)
Thay B vào A ta có :
A = -1-\(\dfrac{24}{25}\)
=> A = \(\dfrac{-49}{25}\)
Cho mik một tick nhé thankss
B\(=-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}\)
\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)
\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)
\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)
\(\dfrac{1}{2}B=\dfrac{-49}{50}\)
\(B=\dfrac{-49}{25}\)
\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=-2*49/50
=-49/25
Tính:
-1- 1/3 - 1/6 - 1/10 - 1/15 - ... - 1/1225
\(=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=-2\cdot\dfrac{49}{50}=\dfrac{-49}{25}\)
\(B= -1-1/3-1/6-1/10-1/15-...-1/1225\)
1+3+6+10+15+...+1225
\(=\dfrac{1\cdot2+2\cdot3+3\cdot4+4\cdot5+...+49\cdot50}{2}\)
\(=\dfrac{49\cdot50\cdot51}{3}=49\cdot50\cdot17=41650\)
Tính giá trị biểu thức:
B= -1 - 1/3 - 1/6 - 1/10 - 1/15 - ... - 1/1225
\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)
\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(B=-2\left(1-\frac{1}{50}\right)\)
\(B=-2\cdot\frac{49}{50}\)
\(B=-\frac{49}{25}\)
C=1+3+6+10+15+...+1225
Tính:
C=1+3+6+10+15+...+1225
\(2C=2+6+12+20+...+2450\)
\(2C=1.2+2.3+3.4+4.5+...+49.50\)
\(6C=1.2.3+2.3.3+3.4.3+...+49.50.3\)\(6C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(6C=1.2.3+2.3.4-1.2.3+...+49.50.51-49.50.51\)
\(6C=49.50.51\)
\(6C=124950\)
\(C=20825\)