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Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
= \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)
= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
A = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)
= \(1-\frac{1}{2^{50}}\)
Vậy \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)= \(1-\frac{1}{2^{50}}\)
a, \(\dfrac{4}{7}\). \(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{1}{21}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{1}{21}\) + \(\dfrac{1}{3}\)
\(\dfrac{4}{7}\).\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\)
\(\dfrac{a}{b}\) = \(\dfrac{8}{21}\):\(\dfrac{4}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\)
b, \(\dfrac{a}{b}\) + \(\dfrac{2}{3}\).\(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) + \(\dfrac{2}{9}\) = \(\dfrac{2}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{3}\) - \(\dfrac{2}{9}\)
\(\dfrac{a}{b}\) = \(\dfrac{4}{9}\)
c, \(\dfrac{a}{b}\) - \(\dfrac{1}{2}.\)\(\dfrac{2}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{2}{7}\) + \(\dfrac{1}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{13}{21}\)
d, \(\dfrac{11}{13}\): \(\dfrac{a}{b}\): \(\dfrac{2}{3}\) = 2\(\dfrac{7}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\):\(\dfrac{2}{3}\) = \(\dfrac{33}{13}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{33}{13}\) \(\times\) \(\dfrac{2}{3}\)
\(\dfrac{11}{13}\): \(\dfrac{a}{b}\) = \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{11}{13}\) : \(\dfrac{66}{39}\)
\(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)
a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4
a, A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2...2 ( 22...2 có 16 số 2)
A = 2 + 22 + 23 + 24 + ... + 216
2A = 22 + 23 + 24 + 25 + ... + 217
2A - A = ( 22 + 23 + 24 + 25 + ... + 217) - ( 2 + 22 + 23 + 24 + ... + 216)
A = 217 - 2
b, B = 1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28
1/2 x B = 1/2 + 1/6 + 1/12 + ... + 1/56
1/2 x B = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/7x8
1/2 x B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8
1/2 x B = 1 - 1/8 = 7/8
B = 7/8 : 1/2
B = 7/8 x 2 = 7/4
\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)
co so cac so la
(22-1,1):(2,2-1,1)+1=20(so)
tong cac so la
(22+1,1)x20:2=231
k nha
Tính: 1.1 + 2.2 + 3.3 + ... + 20.9 + 22
Giải:
Dãy tổng trên là dãy cách đều có các số hạng hơn kém nhau 1.1 đơn vị.
Dãy trên có số số hạng là:
(22 - 1.1) : 1.1 +1 = 20 (số)
Tổng của dãy trên là:
(1.1 + 22) x 20 : 2 = 231
Vậy 1.1 + 2.2 + 3.3 + ... + 20.9 + 22 = 231
\(1\cdot2+2\cdot2+3\cdot2+...+200\cdot2\)
\(=2\left(1+2+3+...+200\right)\)
\(=2\cdot\frac{200\left(200+1\right)}{2}\)
\(=40200\)
áp dụng công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)nha bn!
\(H=\frac{2\cdot2}{1\cdot5}+\frac{2\cdot2}{5\cdot9}+...+\frac{2\cdot2}{45.49}\)
\(H=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{45\cdot49}\)
\(H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(H=1-\frac{1}{49}\)
\(H=\frac{48}{49}\)
\(H=\frac{2.2}{1.5}+\frac{2.2}{5.9}+\frac{2.2}{9.13}+...+\frac{2.2}{45.49}\)
\(\Rightarrow H=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{45.49}\)
\(\Rightarrow H=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{49-45}{45.49}\)
\(\Rightarrow H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(\Rightarrow H=1-\frac{1}{49}=\frac{48}{49}\)