Ta có: \(S=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\)
\(=39\left(1+...+3^{2008}\right)⋮39\)
Ta có: \(S=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{2006}+3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)
\(=121\cdot3\left(1+...+3^{2006}\right)⋮121\)
cho S = \(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+.....+\frac{2010}{2^{2009}}+\frac{2011}{2^{2010}}\)
SO SÁNH S VỚI 3
tính 2S + 3 = 3x
biết S = 3 + 3^2 + 3^3 +...+ 3^2010
giúp mình câu này nhé, mình cảm ơn nhìu
Ai giup minh cau nay duoc k ạ.cảm ơn ạ
2S+3=3^x
biết S=3+3^2+3^3+...+3^2010
Cho A = 1 - 3/4 + (3/4)2 (3/4)3 + (3/4)4 -.......- (3/4)2009 + (3/4)2010
C/m: A không phải là số nguyên
S=\(2+2^2+2^3+...+2^{\text{2010}}\) chia hết cho 7,31
Tinh tong:
a) s1 = 1+2+3+...+999
b) s2 = 10+12+14+...+2010
c) s3 = 21+23+25+...+1001
Tính:\(^{1^2}\)
a) 1^2-2^2+3^2-4^2+5^2-6^2+....+2009^2-2010^2
b) 3^2010-(3^2009+3^2008+3^2007+...+3+3^0)
C/m 3/1*1+2*2+5/2*2+3*3+...+4019/2009*2009+2010*2010<1
Tính (-2) (-3/2) (-4/3)......(-2010/2009) (-2011/2010)
