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Đề sai, đề đúng phải là \(VT< \frac{1}{20}\)
Dễ dàng chứng minh đề sai, ta có:
\(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}>\frac{1}{5^2}+\frac{1}{5^3}=\frac{6}{125}>\frac{1}{24}\)
Còn chứng minh \(VT< \frac{1}{20}\) thì như sau:
\(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)
\(\Rightarrow5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2005}}\)
\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)
\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=A\)
\(\Rightarrow4A=\frac{1}{5}-\frac{1}{5^{2006}}< \frac{1}{5}\)
\(\Rightarrow A< \frac{1}{20}\)
Bài 1:
a) Sửa lại là: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\) nhé.
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.\left(9+1\right)-2^{n-1}.2.\left(4+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)\)
Vì \(10⋮10\) nên \(10.\left(3^n-2^{n-1}\right)⋮10.\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\left(\forall n\in N^X\right).\)
Chúc bạn học tốt!
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)
\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)
Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006
=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)
=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)
=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)
=>A=1/1004+1/1005+.....+1/2006
Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )