Biết 22 +22+23+....+22013=2a . Tìm a
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\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)
\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)
\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)
\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)
1,\(54.\left(23-11\right)+23.\left(-22-54\right)\)
\(=54.12+23.\left(-76\right)\)
\(=648-1748\)
\(=-1100\)
2,\(-7+2x=-17-\left(-36\right)\)
\(=>-7+2x=-17+36\)
\(=>-7+2x=19\)
\(=>2x=19+7=26\)
\(=>x=\frac{26}{2}=13\)
3,\(5⋮2a+3\)
\(=>2a+3\inƯ\left(5\right)\)
\(=>2a+3\in\left\{-1;-5;1;5\right\}\)
\(=>2a\in\left\{-4;-8;-2;2\right\}\)
\(=>a\in\left\{-2;-4;-1;1\right\}\)
Vậy ...
54 . ( 23 - 11 ) + 23 . ( -22 - 54 )
=54.23-54.11+23.-22-23.54
=-54.11+23.-22
=-22(27+23)
=-22.50
=-1100
-7 + 2x = -17 - ( -36 )
-7+2x=-17+36
-7+2x+17-36=0
-26+2x=0
2x=26
x=13
Vậy x=13
Để \(5⋮\left(2a+3\right)\Leftrightarrow\left(2a+3\right)\inƯ\left(5\right)\)
\(\Leftrightarrow\left(2a+3\right)\in\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow a\in\left\{-1;-2;1;-4\right\}\)
Câu 3:
\(A=3+3^2+...+3^{100}\)
\(3A=3^2+3^3+...+3^{101}\)
\(3A-A=3^2+3^3+...+3^{101}-\left(3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-3\)
Mà: \(2A+3=3^N\)
\(\Rightarrow3^{101}-3+3=3^N\)
\(\Rightarrow3^{101}=3^N\)
\(\Rightarrow N=101\)
Vậy: ...
Câu 1:
\(A=4+2^2+...+2^{20}\)
Đặt \(B=2^2+2^3+...+2^{20}\)
=>\(2B=2^3+2^4+...+2^{21}\)
=>\(2B-B=2^3+2^4+...+2^{21}-2^2-2^3-...-2^{20}\)
=>\(B=2^{21}-4\)
=>\(A=B+4=2^{21}-4+4=2^{21}\) là lũy thừa của 2
Câu 6:
Đặt A=1+2+3+...+n
Số số hạng là \(\dfrac{n-1}{1}+1=n-1+1=n\left(số\right)\)
=>\(A=\dfrac{n\left(n+1\right)}{2}\)
=>\(A⋮n+1\)
Câu 5:
\(A=5+5^2+...+5^8\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+\left(5^7+5^8\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+5^6\left(5+5^2\right)\)
\(=30\left(1+5^2+5^4+5^6\right)⋮30\)
\(21+22+23+...+n+4840\)
\(\Rightarrow\left[\left(n-21\right):1+1\right]\left(n+21\right):2=4840\)
\(\Rightarrow\left(n-20\right)\left(n+21\right)=9680\)
\(\Rightarrow n^2+n-420=9680\)
\(\Leftrightarrow n^2+n-100100=0\)
\(\Leftrightarrow n^2-100n+101n-100100=0\)
\(\Leftrightarrow n\left(n-100\right)+101\left(n-100\right)=0\)
\(\Leftrightarrow\left(n+101\right)\left(n-100\right)=0\)
\(\Leftrightarrow\left[n=-101\text{(loại)},n=100\right]\)
\(\Rightarrow n=100\)
\(\text{Hok tốt!}\)
\(\text{@Kaito Kid}\)
21 + 22 + 23 + ... + n = 4840
=> [(n - 21) : 1 + 1](n + 21) : 2 = 4840
=> (n - 20)(n + 21) = 9680
=> n2 + n - 420 = 9680
<=> n2 + n - 10100 = 0
<=> n2 - 100n + 101n - 10100 = 0
<=> n(n - 100) + 101(n - 100) = 0
<=> (n + 101)(n - 100) = 0
<=> \(\orbr{\begin{cases}n=-101\left(\text{loại}\right)\\n=100\end{cases}}\)
Vậy n = 100
ta có: \(S=1-2+2^2-2^3+2^4-2^5+...+2^{2013}-2^{2014}\)
\(\Rightarrow2S=2-2^2+2^3-2^4+2^5-2^6+...+2^{2014}-2^{2015}\)
=> 2S + S = -22015 + 1
=> 3S = -22015 + 1
=> 3S - 1 = -22015
=> 1 - 3S = 22015
( cn về S = 1 - 2 + 22 - 23 + 24-25+...+22013 - 22014 mk vx chưa hiểu quy luật của nó lắm, thật lòng xl bn nha! mk chỉ bk z thoy!)
Các phân số thỏa mãn là: 102/115, 104/115, 106/115, 108/115, 109/115
Nguyễn Ngô Minh Ngọc
ta nhân phan so 22/25 len thanh 66/75 va phan so 22/23 se thanh 66/69
vay 5 phan so do la 66/70,66/71,66/72,66/73,66/74
Đặt A=22+22+23+.....+22013=2a
2A=2(22+22+23+.....+22013)=23+23+24+.....+22014
2A-A=(23+23+24+.....+22014)-(22+22+23+.....+22013)
A=(23-23)+(24-24)+.....+(22013-22013)+(22014-22)+(23-22)
A=22014-4+(8-4)=22014-4+4=22014=2a
suy ra: a=2014