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1/a/ \(A=2+2^2+2^3+....+2^{10}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^9+2^{10}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^9\left(1+2\right)\)
\(=2.3+2^3.3+....+2^9.3\)
\(=3\left(2+2^3+.....+2^9\right)⋮3\)
\(\Leftrightarrow A⋮3\left(đpcm\right)\)
b/ \(A=2+2^2+2^3+....+2^{10}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+2^6.31\)
\(=31\left(2+2^6\right)⋮31\)
\(\Leftrightarrow A⋮31\left(đpcm\right)\)
2/ Với mọi n là số tự nhiên thì \(n\) có hai dạng :
\(\left[{}\begin{matrix}n=2k\\n=2k+1\end{matrix}\right.\)
+) \(n=2k\Leftrightarrow B=\left(n+4\right)\left(n+7\right)=\left(2k+4\right)\left(2k+7\right)\)
Mà \(2k+4⋮2\)
\(\Leftrightarrow\left(2k+4\right)\left(2k+7\right)⋮2\)
\(\Leftrightarrow B\) là số chẵn
+) \(n=2k+1\Leftrightarrow B=\left(n+4\right)\left(n+7\right)=\left(2k+1+4\right)\left(2k+1+7\right)=\left(2k+5\right)\left(2k+8\right)\)
Mà \(2k+8⋮2\)
\(\Leftrightarrow\left(2k+5\right)\left(2k+8\right)⋮2\)
\(\Leftrightarrow B\) là số chẵn
Vậy...
1/
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(A=2.3+2^3.3+2^5.5+...+2^9.3=3.\left(2+2^3+...+2^9\right)\)
Do \(3⋮3\Rightarrow A⋮3\)
\(A=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)
\(A=2.31+2^6.31=31\left(2+2^6\right)\)
Do \(31⋮31\Rightarrow A⋮31\)
2/ \(B=\left(n+4\right)\left(n+7\right)\)
Nếu n chẵn, đặt \(n=2k\Rightarrow B=\left(2k+4\right)\left(2k+7\right)=2\left(k+2\right)\left(2k+7\right)\)
Do 2 chẵn nên B chẵn
Nếu n lẻ, đặt \(n=2k+1\Rightarrow B=\left(2k+5\right)\left(2k+8\right)=2\left(2k+5\right)\left(k+4\right)\)
2 chẵn nên B chẵn
Vậy B luôn chẵn với mọi n
3/ Đề là B(112) hay B(121) bạn?
Bài 1:
a: \(=2^{24}+2^{60}=2^{24}\left(2^{36}+1\right)\)
\(=2^{24}\left(2^4+1\right)\cdot A=17\cdot B⋮17\)
b: \(A=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\cdot\left(2+2^5+...+2^{57}\right)\) chia hết cho 3;5;15
\(A=2\left(1+2+2^2+...+2^{59}\right)⋮2\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow M< 1-\dfrac{1}{n}< 1\)
Vậy \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
Để \(M< 1\), ta phải có điều kiện: \(n\in\) R*. Nếu \(n=0\) thì \(M\) không xác định.
\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=1-\dfrac{1}{n}< 1\)
Vậy \(M< 1\) với \(n\in\) R*.
ta có:1/2!<1
2/3!<1
......
......
2015/2016!<1
=>A=1/2!+2/3!+3/4!+......+2015/2016! luôn luôn <1