Question

tính nhanh 

C=\(\frac{1}{1x2x3}\) +\(\frac{1}{2x3x4}\)+\(\frac{1}{3x4x5}\)+..........+\(\frac{1}{37x38x39}\)

D=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+.....+\(\frac{1}{3^8}\)

Answer 1

\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)

Answer 2

Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)

\(\Rightarrow2D=1-\frac{1}{3^8}\)

\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)

Answer 3

Ta có : 

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{37.38.39}\)

\(2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\)

\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(2C=\frac{1}{2}-\frac{1}{38.39}\)

\(C=\frac{\frac{1}{2}-\frac{1}{38.39}}{2}\)

\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

\(C=\frac{1}{2}.\frac{370}{741}\)

\(C=\frac{185}{741}\)

Answer 4

Ta có : 

\(D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3D-D=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(2D=1-\frac{1}{3^8}\)

\(2D=\frac{3^8-1}{3^8}\)

\(D=\frac{3^8-1}{3^8}:2\)

\(D=\frac{3^8-1}{2.3^8}\)

Vậy \(D=\frac{3^8-1}{3^8}\)

Chúc bạn học tốt ~

Answer 5

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)

\(2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\)

\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(2C=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)

\(C=\frac{185}{741}\)

\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(3D=1+\frac{1}{3}+...+\frac{1}{3^7}\)

\(3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

\(2D=1+\frac{1}{3}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^8}\)

\(2D=1-\frac{1}{3^8}\)

\(D=\frac{1-\frac{1}{3^8}}{2}\)

P/s: có j ko hiểu thì ib

Answer 6

ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt