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\(=3^n\cdot9-2^n\cdot4+3^n-2\)
\(=3^n\left(9+1\right)-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
với n > 1,ta có:
M=3n+2-2n+2+3n-2n
=3n+2+3n-(2n+2+2n)
=3n.(32+1)-2n(22+1)
=3n.10-2n.5=3n.10-2n-1.10
=10.(3n-2n-1) chia hết cho 10 hay M tận cùng là 0(đpcm)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}
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2.
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}\\ =\dfrac{1}{2.2}+\dfrac{1}{3.3}+....+\dfrac{1}{n.n}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{\left(n-1\right).n}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
You k làm đc bài 1 ak -_- làm full cho người ta đi chớ :v
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Rightarrow\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{a+b}{ab}\right)\)
\(\Rightarrow\dfrac{1}{c}=\dfrac{a+b}{2ab}\)
\(\Rightarrow ac+bc=2ab\)
\(\Rightarrow ac+bc-ab=ab\)
\(\Rightarrow ac-ab=ab-bc\)
\(\Rightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\left(đpcm\right)\)