Tính nhanh:
A = 2 - 22 + 23 - 24 + 25 - 26 + ... + 22015
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`#3107`
\(A=1+2^1+2^2+2^3+...+2^{2015}\)
\(2A=2+2^2+2^3+2^4+...+2^{2016}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)
\(A=2+2^2+2^3+2^4+...+2^{2016}-1-2-2^2-2^3-...-2^{2015}\)
\(A=2^{2016}-1\)
Vậy, \(A=2^{2016}-1.\)
\(A=2^0+2^1+2^2+...+2^{2015}\)
\(2\cdot A=2^1+2^2+2^3+...+2^{2016}\)
\(A=2A-A=2^{2016}-2^0\)
\(A=2^{2016}-1\)
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/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$
21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30
= (21 + 30) + (22 + 29) + (23 + 28) + (24 + 27) + (25 + 26)
= 51 + 51 + 51 + 51 + 51
= 51 x 5
= 255
=(21+29)+(22+28)+(23+27)+(24+26)+25+30
=50+50+50+50+25+30
=200+25+30
=225+30
=255
Số số hạng của dãy trên là: (30-20):1+1 = 11 (số)
Tổng trên là: (30+20) x 11:2 = 275
Đáp số: 275
A=(1+2+2^2)+2^3(1+2+2^2)+...+2^2013(1+2+2^2)+2^2016
=7(1+2^3+...+2^2013)+2^2016
Vì 2^2016 chia 7 dư 1
nên A chia 7 dư 1
Tính nhanh
19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 25 + 26
1/3 + 1/4 + 1/5 + 4/6 + 9/12 + 16/20
\(19+18+17+16+14+21+22+23+24+25+26\)
\(=\left(19+21\right)+\left(18+22\right)+\left(17+23\right)+\left(16+24\right)+\left(14+26\right)+25\)
\(=30+30+30+30+30+25\)
\(=175\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{6}+\dfrac{9}{12}+\dfrac{16}{20}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)
\(\text{=}1+1+1\)
\(\text{=}3\)
\(A=2-2^2+2^3-2^4+......+2^{2015}\)
\(2A=2^2-2^3+2^4-2^5+.....+2^{2016}\)
\(2A+A=2^2-2^3+2^4-2^5+.....+2^{2016}+\left(2-2^2+2^3-2^4+.....+2^{2015}\right)\)
\(3A=2^{2016}+2\)
\(\Rightarrow A=\frac{2^{2016}+2}{3}\)
Ta có :
\(A=2-2^2+2^3-2^4+...+2^{2015}\)
\(\Leftrightarrow\)\(A=\left(-2^2-2^4-...-2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)\)
\(\Leftrightarrow\)\(A=-\left(2^2+2^4+...+2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)\)
Gọi \(M=2^2+2^4+...+2^{2014}\)
\(\Leftrightarrow\)\(4M=2^4+2^6+...+2^{2016}\)
\(\Leftrightarrow\)\(4M-M=\left(2^4+2^6+...+2^{2016}\right)-\left(2^2+2^4+...+2^{2014}\right)\)
\(\Leftrightarrow\)\(3M=2^{2016}-2^2\)
\(\Leftrightarrow\)\(M=\frac{2^{2016}-4}{3}\)
Gọi \(N=2+2^3+...+2^{2015}\)
\(\Leftrightarrow\)\(4N=2^3+2^5+...+2^{2017}\)
\(\Leftrightarrow\)\(4N-N=\left(2^3+2^5+...+2^{2017}\right)-\left(2+2^3+...+2^{2015}\right)\)
\(\Leftrightarrow\)\(3N=2^{2017}-2\)
\(\Leftrightarrow\)\(N=\frac{2^{2017}-2}{3}\)
\(\Rightarrow\)\(A=-\left(2^2+2^4+...+2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)=-\frac{2^{2016}-4}{3}+\frac{2^{2017}-2}{3}\)
\(\Rightarrow\)\(A=\frac{\left(-1\right).\left(2^{2016}\right)+2^{2017}.1+4-2}{3}=\frac{2^{2016}\left(2-1\right)+2}{3}=\frac{2^{2016}+2}{3}\)
Vậy \(A=\frac{2^{2016}+2}{3}\)