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Bài 1:
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow\dfrac{1}{5}A=\dfrac{1}{5^2}+\dfrac{1}{5^3}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{100}}\)
Lây vế trừ vế, ta được:
\(A-\dfrac{1}{5}A=\dfrac{4}{5}A\)
\(\dfrac{4}{5}A=\dfrac{1}{5}-\dfrac{1}{5^{100}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{5}-\dfrac{1}{5^{100}}}{\dfrac{4}{5}}=\dfrac{\dfrac{1}{5}.\left(1-\dfrac{1}{5^{99}}\right)}{\dfrac{1}{5}.4}=\dfrac{1-\dfrac{1}{5^{99}}}{4}\)
Vậy \(A=\dfrac{1-\dfrac{1}{5^{99}}}{4}\).
Chúc bạn học tốt!
Bài 2:
Có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\)
\(\Leftrightarrow B=273+...+3^{1986}.273\)
\(\Leftrightarrow B=273\left(1+...+1986\right)\)
Vì \(273⋮13\)
Nên \(B=273\left(1+...+1986\right)⋮13\)
Vậy \(B⋮13\)
Lại có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+3^{1984}\left(3+3^3+3^5+3^7\right)\)
\(\Leftrightarrow B=2460+...+3^{1984}.2460\)
\(\Leftrightarrow B=2460\left(1+...+3^{1984}\right)\)
Vì \(2460⋮41\)
Nên \(B=2460\left(1+...+3^{1984}\right)⋮41\)
Vậy \(B⋮41\).
Chúc bạn học tốt!
tính M hay chứng minh M ko là stn hay đầu bài là j vậy bn????
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
a) \(S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}\)
\(\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2\) vậy \(S=-2\)
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự