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5 + 52 + 53 + ... + 599
= 5.(1 + 5 + 52) + 54.(1 + 5 + 52) + ... + 597.(1 + 5 + 52)
= 5.31 + 54.31 + ... + 597.31
= 31.(5 + 54 + .. + 597) chia hết cho 31
4 + 42 + 43 + ... + 499
= 4.(1 + 4 + 42) + 44.(1 + 4 + 42) + ... + 497.(1 + 4 + 42)
= 4.21 + 44.21 + ... + 497.21
= 4.21.(1 + 43 + ... + 496)
= 4.7.3.(1 + 43 + ... + 496)
= 28.3.(1 + 43 + ... + 496) chia hết cho 28
A = 21+22+23+24+...+2100
A = (21+22)+(23+24)+...+(299+2100)
A = 2(1+2) + 23(1+2) +....+ 299.(1+2)
A = 2.3 + 23.3 +....+ 299.3
A = 3.(2+23+...+299) chia hết cho 3
=> A chia hết cho 3 (đpcm)
A = 21+22+23+24+...+2100
A = (21+22+23+24)+(25+26+27+28)+...+(297+298+299+2100)
A = 2(1+2+22+23)+25(1+2+22+23)+...+297(1+2+22+23)
A = 2.15 + 25.15 +....+ 297.15
A = 15.(2+25+...+297) chia hết cho 5 (vì 15 chia hết cho 5)
=> A chia hết cho 5 (Đpcm)
B = 3x+3 + 3x+1 + 2x+3 + 2x+2
= 3x.33 + 3x.3 + 2x.23 + 2x.22
= 3x(33 + 3) + 2x(23 + 22)
= 3x.30 + 2x.12
Vì 3x.30 chia hết cho 6 => 3x.30 + 2x.12 chia hết cho 6
2x.12 chia hết cho 6
=> B chia hết cho 6
a) Đặt biểu thức trên là A, ta có:
A = 21 + 22 + 23 + 24 + ... + 299 + 2100
=> A = (21 + 22) + (23 + 24) + ... + (299 + 2100)
=> A = 21.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2)
=> A = 21.3 + 23.3 + ... + 299.3
=> A = 3(21 + 23 + ... + 299)
=> A ⋮ 3
\(26=13.2\)
\(s=3.\left(1+3+9\right)+3^4.\left(1+3+9\right)+....+3^{2012}.\left(1+3+9\right)\)
\(s=3.13+3^413+.....+3^{2012}.13\)
\(s=13.\left(3+3^4+....+3^{2012}\right)\)
\(\Rightarrow s=3.\left(1+3\right)+3^3.\left(1+3\right)+.......+3^{2015}.\left(1+3\right)\)
\(s=3.4+3^3.4+....+3^{2015}.4\)
\(s=4.\left(3+3^3+.....+3^{2015}\right)\)
\(\Rightarrow4⋮2\Rightarrow4.\left(3+3^3+....+3^{2015}\right)⋮2\)
\(\Rightarrow s⋮2\Leftrightarrow s⋮13\)
\(\Rightarrow s⋮\orbr{\begin{cases}13\\2\end{cases}}\Leftrightarrow s⋮26\)
Bài 1 : Ta có : \(A=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)\)
\(=\overline{......0}\)
\(\Rightarrow\)Chữ số tận cùng của \(A\)là \(0\)
Bài 3:
a)Ta có : \(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+...+2^{96}\left(2+2^2+2^3+2^4\right)\)
\(=31+2^4.31+...+2^{96}.31\)
\(=31\left(1+2^4+...+2^{96}\right)⋮31\)
\(\Rightarrow\)\(đpcm\)
b) Ta có : \(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(\Rightarrow2C=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow2C-C=\left(2^2+2^3+2^4+...+2^{100}+2^{101}\right)-\left(2+2^2+2^3+...+2^{99}+2^{100}\right)\)
\(\Rightarrow C=2^{101}-2\)
Mà \(2^{2x}-2=C\)
\(\Rightarrow2^{2x}-2=2^{101}-2\)
\(\Rightarrow2^{2x}=2^{101}\)
\(\Rightarrow2x=101\)
\(\Rightarrow x=\frac{101}{2}\)
Vậy \(x=\frac{101}{2}\)
Bài 2:
Ta có : \(\overline{abcd}=1000a+100b+10c+d\)
\(=1000a+96b+8c+\left(d+2c+4b\right)\)
\(=8\left(125a+12b+c\right)+\left(d+2c+4b\right)\)
Vì \(\hept{\begin{cases}d+2c+4b⋮8\\8\left(125a+12b+c\right)⋮8\end{cases}}\)
\(\Rightarrow\overline{abcd}⋮8\)
\(\Rightarrowđpcm\)
nó chia hết cho 31 vì 2 mũ 100+2 chia hết cho 31
A=\(\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{2011}+2^{2012}+2^{2013}+2^{2014}+2^{2015}\right)\)
A=\(2\left(1+2+2^2+2^3+2^4\right)+...+2^{2011}\left(1+2+2^2+2^3+2^4\right)\)
A= \(2.31+...+2^{2011}.31\)
=> \(A⋮31\)
chuc ban hoc tot