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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)
A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).0}{1-2+3-4+...+99-100}\)
A= 0
KẾT QUẢ ĐÚNG 100%
Ta thấy : \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
nên kết quả dãy trên bằng 0
\(A=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(75,6-75,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot0}{1-2+3-4+...+99-100}\)
\(=0\)
Đặt \(E=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\)
\(\Rightarrow7E=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\)
\(\Rightarrow7E-E=\left(1+\frac{1}{7}+...+\frac{1}{7^{98}}+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6E=1-\frac{1}{7^{100}}\)
\(\Rightarrow E=\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=\left(36-\frac{36}{7^{100}}\right):\frac{1-\frac{1}{7^{100}}}{6}\)
\(\Rightarrow A=36\left(1-\frac{1}{7^{100}}\right).\frac{6}{1-\frac{1}{7^{100}}}\)
\(\Rightarrow A=36.6=216\)