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Bài 1:
b) Ta có:
\(16^5=2^{20}\)
\(\Rightarrow B=16^5+2^{15}=2^{20}+2^{15}\)
\(\Rightarrow B=2^{15}.2^5+2^{15}\)
\(\Rightarrow B=2^{15}\left(2^5+1\right)\)
\(\Rightarrow B=2^{15}.33\)
\(\Rightarrow B⋮33\) (Đpcm)
c) \(C=5+5^2+5^3+5^4+...+5^{100}\)
\(\Rightarrow C=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow C=1\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)
\(\Rightarrow\left(1+5^2+...+5^{98}\right)\left(5+5^2\right)\)
\(\Rightarrow C=Q.30\)
\(\Rightarrow C⋮30\) (Đpcm)
Bài 1 : a, \(A=1+3+3^2+...+3^{118}+3^{119}\)
\(A=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(A=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(A=1.30+...+3^{116}.30=\left(1+...+3^{116}\right).30⋮3\)
Vậy \(A⋮3\)
b, \(B=16^5+2^{15}=\left(2.8\right)^5+2^{15}\)
\(=2^5.8^5+2^{15}=2^5.\left(2^3\right)^5+2^{15}\)
\(=2^5.2^{15}+2^{15}.1=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
Vậy \(B⋮33\)
c, Tương tự câu a nhưng nhóm 2 số
Bài 2 : a, \(n+2⋮n-1\) ; Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+2\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+2-n+1⋮n-1\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\in\left\{1;3\right\}\Rightarrow n\in\left\{2;4\right\}\)
Vậy \(n\in\left\{2;4\right\}\) thỏa mãn đề bài
b, \(2n+7⋮n+1\)
Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\)
\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\Rightarrow n\in\left\{0;4\right\}\)
Vậy \(n\in\left\{0;4\right\}\) thỏa mãn đề bài
c, tương tự phần b
d, Vì : \(4n+3⋮2n+6\)
Mà : \(2n+6⋮2n+6\Rightarrow2\left(2n+6\right)⋮2n+6\Rightarrow4n+12⋮2n+6\)
\(\Rightarrow\left(4n+12\right)-\left(4n+3\right)⋮2n+6\)
\(\Rightarrow4n+12-4n-3⋮2n+6\Rightarrow9⋮2n+6\)
\(\Rightarrow2n+6\in\left\{1;2;9\right\}\Rightarrow2n=3\Rightarrow n\in\varnothing\)
Vậy \(n\in\varnothing\)
Đặt A=\(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
3A=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{98}}\)
2A = 3A - A = \(\frac{1}{3}-\frac{1}{3^{98}}\)<\(\frac{1}{2}\)
=> A = \(\frac{\frac{1}{3}-\frac{1}{3^{98}}}{2}<\frac{1}{2}\)(đpcm)
Đặt B=2100-299+298-297+.....+22-2
2A=2101-2100+299-298+...+23-22
3A=2A+A=2101-2
=> A=\(\frac{2^{101}-2}{3}\)