Chứng mình rằng:
S = 2 + 2 ^ 2 + 2 ^ 3 + .... + 2 ^ 8 ⋮ -6
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Ta có :
S = 5 + 52 +53 +54 +.... + 5100 có (100 - 1) : 1 + 1 = 100 số hạng
S = (5 + 52) + (53 + 54) + ....... + (599 + 5100)
S = 5 . (1 + 5) + 53 . (1 + 5) + .... + 599 . (1 + 5)
S = 5 . 6 + 53 . 6 + ..... + 599 . 6
S = 6 . (5 + 53 + ..... + 599)
Vì 6 chia hết cho 6 nên S chia hết cho 6 (ĐPCM)
Ủng hộ mk nha !! ^_^
\(S=5+5^2+5^3+...+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+.....+\left(5^{99}+5^{100}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+......+5^{99}\left(1+5\right)\)
\(=\left(1+5\right)\left(5+5^3+.....+5^{99}\right)\)
\(=6\left(5+5^3+....+5^{99}\right)\)
1:
Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)
2) Để A là số nguyên dương thì
\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{6;12\right\}\)
S=(1+2)+(2^2+2^3)+(2^4+2^5)+....+(2^99+2^100)
S=3+3.2^2+3.2^4+.....+3.2^99
S=3.(2^2+2^4+.....+2^99)
Vì 3 chia hết 3=>3.(2^2+2^4+....+2^99)
=>S chia hết 3
2S=2+2^2+2^3+2^4+.....+2^101
2S-S=(2+2^2+2^3+2^4+....+2^101)-(1+2+2^2+2^3+2^4+....+2^100)
S=2^101-1
S+1=2^101-1+1=2^101
=>x=101
S = (21+22)+(23+24)+...+(299+2100)
S = 2.(1+2)+23.(1+2)+...+299.(1+2)
S = 2.3+23.3+...+299.3
S = 3.(2+23+...+299)
=> S chia hết cho 3
S = (21+22+23+24)+(25+26+27+28)+...+(297+298+299+2100)
S = 2.(1+2+4+16)+25.(1+2+4+16)+...+297.(1+2+4+16)
S = 2.15+25.15+...+297.15
S = 15.(2+25+...+297)
=> S chia hết cho 15
Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99
=2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
=1+0+0+0+...+0+1/99
=1+1/99
=100/99
Mà 100/99>1.Suy ra S>1
Vậy S>1
Có 2+2^2+2^3+....+2^8
=(2+2^2)+(2^3+2^4)+.....+(2^7+2^8)
=(2+4)+2^2(2+2^2)+.....+2^6(2+2^2)
=6+2^2(2+4)+......+2^6(2+4)
=1.6+2^2.6+....+2^6.6
=6(1+2^2+....+2^6)
Vì 6 chia hết cho -6 ; 1+2^2+...+2^6 thuộc Z
=>6(1+2^2+....+2^6) chia hết cho -6
hay 2+2^2+2^3+....+2^8 chia hết cho -6