Tính \(A=1+10+10^2+10^3+.....+10^{100}\)
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a.
\(A=1+3+3^2+3^3+...+3^n\)
\(3A=3+3^2+3^3+3^4+...+3^{n+1}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{n+1}\right)-\left(1+3+3^2+3^3+...+3^n\right)\)
\(2A=3^{n+1}-1\)
\(A=\frac{3^{n+1}-1}{2}\)
b.
\(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\)
\(10B=10+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^{98}}+\frac{1}{10^{99}}\)
\(10B-B=\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{99}}+\frac{1}{10^{100}}\right)-\left(10+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^{98}}+\frac{1}{10^{99}}\right)\)
\(9B=\frac{1}{10^{100}}-10\)
\(B=\frac{\frac{1}{10^{100}}-10}{9}\)
A = 1 - 2 + 3 - 4 + 5 - 6 + ... + 99 - 100
Biểu thức A có : (100 - 1) : 1 + 1 = 100 (số hạng)
Nhóm hai số hạng thành 1 nhóm ta được : 100 : 2 = 50 (nhóm)
=> A = (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100)
=> A = (-1) + (-1) + (-1) + ... + (-1) (50 thừa số -1)
=> A = -1 . 50
=> A = -50
Vậy A = -50
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Bài 1 :
a) = 98765
b) = 300657
c) 610000
Bài 2:
a) X + 77 = 1800
X = 1800 - 77
X = 1723
b) 8950 : x = 5
x = 8950 : 5
x = 1790
c) x = 8765 : 5
x = 1753
a,98765 b,300657 c,610000 Bài 2 : a, 17923 b, 1790 c, 1753
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1-\frac{1}{3^{100}}}{2}\)
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)
\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)
\(B=\frac{15}{14}:3=\frac{5}{14}\)
a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\frac{3}{14}\)
\(\Rightarrow B=\frac{5}{14}\)
A = 1 + 10 + 102 + ... + 10100
10A - A = ( 10 + 102+ 103 + ... + 10101) - ( 1 + 10 + 102+ ... + 10100 )
9A = 10101 - 1
=> A = 10101 - 1/9
\(A=1+10+10^2+10^3+......+10^{100}\)
\(10A=10+10^2+10^3+.....+10^{101}\)
\(10-A=10^{101}-1\)
\(9A=10^{101}-1=>A=\frac{10^{101}-1}{9}\)