Tính tổng sau:
\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
Những câu hỏi liên quan
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 = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
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Tính các tổng sau
\(A=7-7^4+7^7-...+7^{301}\)
\(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(C=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
AI TÍNH ĐƯỢC MÌNH CHO 10 TICK NHA
Tính tổng sau
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^50}\)
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^5}\)
\(\Rightarrow7A=1+\frac{1}{7}+...+\frac{1}{7^4}\)
\(\Rightarrow7A-A=1-\frac{1}{7^5}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^5}}{6}\)
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a) Tính tổng: \(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
b) Chứng minh rằng : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}
a)S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007
=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006
=>(7-1)S=6-(1/7)^2007
=>S=1-(-1/7^2007/6)
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Tính gọn tổng sau :
a) \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...\left(-\frac{1}{7}\right)^{2007}\)
b)\(Q=\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
giúp mình nha các bạn, cam on nhìu
Tính \(A=\left(36-\frac{36}{7^{100}}\right):\left(\frac{1}{7^1}+\frac{1}{7^2}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\right)\)
Đặt \(E=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\)
\(\Rightarrow7E=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\)
\(\Rightarrow7E-E=\left(1+\frac{1}{7}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6E=1-\frac{1}{7^{100}}\)
\(\Rightarrow E=\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=\left(36-\frac{36}{7^{100}}\right):\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=36\left(1-\frac{1}{7^{100}}\right).\frac{6}{1-\frac{1}{7^{100}}}\)
\(\Rightarrow A=36.6=216\)
thực hiên các phép tính tính :
a) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
b) \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Tính
A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
B = \(\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\frac{A}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\)
\(A-\frac{A}{7}=\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\right)\)
\(\frac{6}{7}A=\frac{1}{7}-\frac{1}{7^{101}}\)
\(A=\left(\frac{1}{7}-\frac{1}{7^{101}}\right).\frac{7}{6}\)
\(A=\frac{1}{6}-\frac{1}{6.7^{100}}\)
\(B=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
\(=4.\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)\)
Gọi \(C=\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\)
\(\frac{C}{5}=\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\)
\(C-\frac{5}{C}=\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\right)\)
\(\frac{4}{5}C=\frac{1}{5}-\frac{1}{5^{201}}\)
\(C=\left(\frac{1}{5}-\frac{1}{5^{201}}\right).\frac{5}{4}\)
\(=\frac{1}{4}-\frac{1}{4.5^{200}}\)
Thay vào B ta có
\(B=4.\left(\frac{1}{4}-\frac{1}{4.5^{200}}\right)\)
=\(=1-\frac{1}{5^{200}}\)
Tinh cac tong sau
\(H=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(I=\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+...+\frac{4}{5^{200}}\)