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\(A=1+2+2^2+2^3+...+2^{10}\)
\(2A=2+2^2+2^3+2^4+...+2^{11}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)
\(A=2^{11}-1< 2^{11}\)
\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)
\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)
\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)
\(B=-2^{11}+2^3+10.2^{11}-2^3\)
\(B=9.2^{11}\)
Ta cần so sánh: \(9.2^{11}\) và \(2^{14}\)
Hay \(9\) và \(2^3\)
\(9>8=2^3\Leftrightarrow B>2^{14}\)
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Bài làm
Đặt a - b = x ; b - c = y ; c - a = z
=> x + y + z = 0
Ta có :
\(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{x+y+z}{xyz}\right)\)
=> \(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)( Vì x + y + z = 0 )
Vậy ta có đpcm
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A = 2.22 + 3.23 + 4.24 + ... + n.2n
2.A = 2.23 + 3.24 + 4.25 + ...+ n.2n+1
=> A - 2.A = 2.22 + (3.23 - 2.23) + (4.24 - 3.24) + ...+ (n - n + 1).2n - n.2n+1
=> A = 2.22 + 23 + 24 + ..+ 2n - n.2n+ 1 = 22 + (22 + 23 + ....+ 2n+ 1) - (n+1).2n+1
=> A = - 22 - (22 + 23 + ....+ 2n+ 1) + (n+1).2n+1
Tính B = 22 + 23 + ....+ 2n+ 1 => 2.B = 23 + ....+ 2n+ 1 + 2n+2 => 2B - B = 2n+2 - 22 => B = 2n+2 - 22
Vậy A = 22 - 2n+2 + 22 + (n+1).2n+1 = (n+1).2n+1 - 2n+ 2 = 2n+1.(n + 1 - 2) = (n-1).2n+1 = 2(n-1).2n
Theo bài cho A = 2(n-1).2n = 2n+10 => 2(n - 1) = 210 => n - 1 = 29 = 512 => n = 513
Vậy.............
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