1+2+22+23+...+22012
2014-2
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a) 23 + (-77) + (-23) + 77 =
[23 + (-23)] + [(-77) + 77]
= …0+0=0……
b) (-2 020) + 2 021 + 21 + (-22)
=[(-2 020) + 2 021] + [21 + (-22)]
= …1……+ (-1)……..
= 0.
a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)
= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)
= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)
= 1 - 1 + \(\dfrac{2022}{2023}\)
= \(\dfrac{2022}{2023}\)
b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)
= \(\dfrac{33}{11}\)
= 3
c, 2000 + { 20 - [ 4.20220 - (32 + 5):2] }
= 2000 + { 20 - [ 4.1 - (9+5):2]}
= 2000 + { 20 - [ 4 - 14 : 2 ]}
= 2000 + { 20 - [ 4 -7]}
= 2000 + { 20 - (-3)}
= 2000 + 23
= 2023
=\(\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{1}{22}-\dfrac{23}{22}\right)+\dfrac{2}{3}\)
=\(\dfrac{17}{17}-\dfrac{22}{22}+\dfrac{2}{3}\)
=\(1-1+\dfrac{2}{3}\)
=0+\(\dfrac{2}{3}\)
=\(\dfrac{2}{3}\)
`1+2+2^2+2^3+....+2^63`
`=2+2+2^2+2^3+....+2^63-1`
`=2.2+2^2+2^3+....+2^63-1`
`=2^2+2^2+2^3+....+2^63-1`
`=2.2^2+2^3+....+2^63-1`
`=2^3+2^3+...2^63-1`
`=2.2^3+....+2^63-1`
`=2^4+....+2^63-1`
`=2^{63}.2-1=2^64-1`
\(2S=2+2^2+...+2^{2022}\\ \Leftrightarrow2S-S=S=2^{2022}-1\)
\(1+2+2^2+2^3+...+2^n=357680\)
\(\Leftrightarrow2\cdot\left(1+2+2^2+...+2^n\right)=2\cdot357680\)
\(\Leftrightarrow2+2^2+2^3+2^4+...+2^{n+1}=2\cdot357680\)
\(\Leftrightarrow\left(2+2^2+...+2^{n+1}\right)-\left(1+2+2^2+...+2^n\right)=2\cdot357680-357680\)
\(\Leftrightarrow\left(2-2\right)+\left(2^2-2^2\right)+...+\left(2^n-2^n\right)+\left(2^{n+1}-1\right)=357680\)
\(\Leftrightarrow2^{n+1}-1=357680\)
\(\Leftrightarrow2^{n+1}=357681\)
Xem lại đề
\(1+2+2^2+2^3+...+2^n=357680\)
\(\Rightarrow\dfrac{2^{n+1}-1}{2-1}=357680\)
\(\Rightarrow2^{n+1}=357680+1\)
\(\Rightarrow2^{n+1}=357681\Rightarrow n+1=\sqrt[]{357681}\Rightarrow n=\sqrt[]{357681}-1\)
`#3107.101107`
Đặt $A = 1 + 2 + 2^2 + 2^3 + ... + 2^{50}$
$2A = 2 + 2^2 + 2^3 + ... + 2^{51}$
$2A - A = (2 + 2^2 + 2^3 + ... + 2^{51}) - (1 + 2 + 2^2 + ... + 2^{50})$
$A = 2 + 2^2 + 2^3 + ... + 2^{51] - 1 - 2 - 2^2 - ... - 2^{50}$
$A = 2^{51} - 1$
Vậy, `A =` $2^{51} - 1.$
Đề đungs
1+22+23+...+22012/2014-2