Tính tổng \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
Tính tổng A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}} \)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\left(1\right)\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\left(2\right)\)
Lấy (2) - (1) ta được:\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow A=\left(\frac{3^{100}-1}{3^{100}}\right):2\)
\(\Leftrightarrow A=\frac{3^{100}-1}{2.3^{100}}\)
tính tổng
A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)(1)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)(2)
Lấy (2) trừ đi (1) ta có :
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\frac{\left(1-\frac{1}{3^{100}}\right)}{2}\)
Tính tổng: A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
ta có 3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)
Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
Tính tổng :
1, A = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.................+\frac{1}{100}\)
2, B = \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....................+\frac{99}{100}\)
Câu 1: Tính tổng \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
3A = 1 + 1/3 + 1/3^2 + ... + 1/3^199
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^100 )
2A = 1 - 1/3^100
A = ( 1 - 1/3^100 ) / 2
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{3^{100}-1}{3^{100}.2}\)
mk chỉ làm được đến đây thôi
Tính nhanh tổng A :
A =\(3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)
\(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)
\(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)
\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6.\left(1-\frac{1}{101}\right)\)
\(=6.\frac{100}{101}=\frac{600}{101}\)
Vậy \(A=\frac{600}{101}\)
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)
\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)
Vay A = ........
Tính tổng
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\)
\(B=\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{2015}{2016}\)
A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100
Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100
A= 99/100
Cảm ơn bạn Kudo Shinichi, nhưng
1=2-1 ->ok
1/2=1-1/2 ->ok
1/3=1/2-1/3 -> sai
vì 1/2-1/3=1/6
Tính tổng S =\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)