Rút gọn biểu thức:C=1+2+22+23+....+299
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1/
Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.
Số số hạng: $(101-1):4+1=26$
$A=(101+1)\times 26:2=1326$
2/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$
a) \(A=1+2+2^2+2^3+...+2^{99}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{100}-1-2-2^2-...-2^{99}=2^{100}-1\)
b) \(A=1+2+2^2+...+2^{99}=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)
\(=15+2^4.15+...+2^{96}.15=15\left(1+2^4+...+2^{96}\right)\)
\(=3.5\left(1+2^4+...2^{96}\right)\) chia hết cho 3 và 5
c) \(A=1+2+2^2+...+2^{99}\)
\(=1+2\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)
\(=1+2.7+...+2^{97}.7=1+7\left(2+...+2^{97}\right)\) chia 7 dư 1
=> A không chia hết cho 7
\(C=\dfrac{\sqrt{y^3}-1}{y+\sqrt{y}+1}-\dfrac{y+3\sqrt{y}+2}{\sqrt{y}+1}\)
\(C=\dfrac{\sqrt{y^3}-1^3}{y+\sqrt{y}+1}-\dfrac{y+\sqrt{y}+2\sqrt{y}+2}{\sqrt{y}+1}\)
\(C=\dfrac{\left(\sqrt{y}+1\right)\left[\left(\sqrt{y}\right)^2+\sqrt{y}\cdot1+1\right]}{y+\sqrt{y}+1}-\dfrac{\left(\sqrt{y}+1\right)\left(\sqrt{y}+2\right)}{\sqrt{y}+1}\)
\(C=\dfrac{\left(\sqrt{y}+1\right)\left(y+\sqrt{y}+1\right)}{y+\sqrt{y}+1}-\left(\sqrt{y}+2\right)\)
\(C=\sqrt{y}+1-\sqrt{y}-2\)
\(C=-3\)
\(C=\dfrac{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}{y+\sqrt{y}+1}-\dfrac{\left(\sqrt{y}+1\right)\left(\sqrt{y}+2\right)}{\sqrt{y}+1}\)
\(=\sqrt{y}-1-\sqrt{y}-2=-3\)
Ta có:
x3(x+2) – x(x3 + 23) – 2x(x2 – 22)
= x3 . x + x3 . 2 – (x . x3 + x . 23) – ( 2x . x2 – 2x . 22)
= x4 + 2x3 – (x4 + 8x ) – (2x3 – 8x)
= x4 + 2x3 – x4 – 8x – 2x3 + 8x
= (x4 – x4) + (2x3 – 2x3) + (-8x + 8x)
= 0
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)
\(\Leftrightarrow2P=2^2+2^3+...+2^{100}\\ \Leftrightarrow2P-P=2^2+2^3+...+2^{100}-2-2^2-...-2^{99}\\ \Leftrightarrow P=2^{100}-2\)
C=1+2+22+23+....+299
=> 2C=2+22+23+....+2100
=> (2C-C)=C=(2+22+23+....+2100)-(1+2+22+23+....+299)
=> C=2100-1