A = 4 2^2 2^3 2^4 ... 2^20
ai bt giup minh voi:v,minh can gap
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a) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=\left(\frac{1}{2}\times2\right)+\left(\frac{1}{4}\times2\right)+\left(\frac{1}{8}\times2\right)+\left(\frac{1}{16}\times2\right)+\left(\frac{1}{32}\times2\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
Ta lấy : \(2A-1A=1A\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}\)
\(A=\frac{31}{32}\)
Vậy \(A=\frac{31}{32}\)
b) Đặt \(B=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)
\(B=2\times(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20})\)
\(B=2\times\left(1-\frac{1}{20}\right)\)
\(B=2\times\frac{19}{20}\)
\(B=\frac{19}{10}\)
Vậy \(B=\frac{19}{10}\)
Học tốt # ^-<
( 20 . 24 + 12 . 24 - 48 . 22 ) : 26
= (5.22. 24 + 3.22 . 24 - 3. 24 . 22):26
= ( 5. 26 +3.26- 3.26) :26
= 26( 5+3-3):26
= 26.5:26
= 5
( 20 . 24 + 12 . 24 - 48 . 22 ) : 26
= [24 .(20 + 12) - 48 . 22] : 26
= [24 . 32 - 48 . 22] : 26
= [22 . 22 . 25 - 48 . 22] : 26
= [22 . 27 - 48 . 22] : 26
= [22 . (27 - 48)] : 26
= [22 . (128 - 48)] : 26
= [4 . 80] : 64
= 320 : 64
= 5
\(A=7+7^2+7^3+........+7^{2016}\)
\(A=7\left(1+7+7^2+7^3+........+7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\)
\(A=7\left[\left(1+7+7^2+7^3\right)+........+\left(7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\right]\)
\(A=7\left[\left(1+7+7^2+7^3\right)+........+7^{2012}\left(1+7+7^2+7^3\right)\right]\)
\(A=7\left[400+........+7^{2012}.400\right]\)
\(A=7.400\left(1+7^4+7^8+7^{12}+......+7^{2012}\right)⋮400\)
Vì \(20^2=400\) nên \(A⋮20^2\left(dpcm\right)\)
Lần sau viết cái đề rõ rõ ra nhs!!!
a) \(A=2+2^2+2^3+................+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
b) \(B=1+3+3^2+..................+3^{2009}\)
\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)
\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)
\(\Rightarrow2B=3^{2010}-1\)
\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)
c) \(C=4+4^2+4^3+................+4^n\)
\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
S1 = 1-2+3-4+....+2017-2018
= (-1)+(-1)+....+(-1)
= (-1) x 1009
= -1009
cái khoảng trống là jv