CHO M =1+1/2+1/3+............+1/2100-1
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1 + 1 + 2 + 2 + 3 + 3 x 100 - 2100
= (1 x 2) + (2 x 2) + (3 x 2) x 100 - 2100
= 2 + 4 + 6 x 100 - 2100
= 6 + 6 x 100 - 2100
= 12 x 100 - 2100
= 1200 - 2100
= -900
Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$
$=2.3+2^3.3+...+2^{99}.3$
$=3(2+2^3+...+2^{99})\vdots 3$
Ta có đpcm.
\(\left(3x-5\right)^{2018}+\left(y^2-1\right)^{2006}+\left(x-z\right)^{2100}=0\)
ta có \(\left\{{}\begin{matrix}\left(x-z\right)^{2100}\ge0\\\left(y^2-1\right)^{2006}\ge0\\\left(3x-5\right)^{2018}\ge0\end{matrix}\right.\)
dấu = xảy ra khi \(\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\z-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\z=x\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=1\\z=\dfrac{5}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-1\\z=\dfrac{5}{3}\end{matrix}\right.\end{matrix}\right.\)
vậy.................
a) \(S=1+2+2^2+...+2^{100}\)
\(2S=2+2^2+2^3+...+2^{101}\)
\(2S-S=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)
\(S=2^{101}-1\)
b) \(X=2^{2012}-2^{2011}-...-2-1\)
\(X=2^{2012}-\left(1+2+...+2^{2011}\right)\)
Đặt \(X=2^{2012}-Y\)
Ta có :
\(Y=1+2+...+2^{2011}\)
\(2Y=2+2^2+...+2^{2012}\)
\(2Y-Y=\left(2+2^2+...+2^{2012}\right)-\left(1+2+...+2^{2011}\right)\)
\(Y=2^{2012}-1\)
\(\Rightarrow X=2^{2012}-2^{2012}+1\)
\(\Rightarrow X=1\)
\(\Rightarrow2010X=2010\)
A = 1 + 3 + 32 + 33 + ... + 3100
3A = 3 + 32 + 33 +34+ .... + 3101
3A - A = (3 + 32 + 34 + ... + 3101) - (1 + 3 + 32 + 33 + ... + 3100)
2A = 3 + 32 + 34 + ... + 3101 - 1 - 3 - 32 - 33 - ... - 3100
2A = (3 - 3) + (32 - 32) + ... + (3100 - 3100) + (3101 - 1)
2A = 3101 - 1
A = \(\dfrac{3^{101}-1}{2}\)
\(A.x=x+x^2+x^3+...+x^{101}\)
\(A.x-A=x^{101}-1\Rightarrow A\left(x-1\right)=x^{101}-1\)
\(\Rightarrow A=\dfrac{x^{101}-1}{x-1}\)