Tính A = 20 + 23 + 25 + 27 +.....+ 299
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`A=2^{0}+2^{1}+2^{2}+....+2^{99}`
`=(1+2+2^{2}+2^{3}+2^{4})+(2^{5}+2^{6}+2^{7}+2^{8}+2^{9})+......+(2^{95}+2^{96}+2^{97}+2^{97}+2^{99})`
`=(1+2+2^{2}+2^{3}+2^{4})+2^{5}(1+2+2^{2}+2^{3}+2^{4})+.....+2^{95}(1+2+2^{2}+2^{3}+2^{4})`
`=31+2^{5}.31+....+2^{95}.31`
`=31(1+2^{5}+....+2^{95})\vdots 31`
\(A=2^0+2^1+2^2+2^3+2^4+2^5+2^6+...+2^{99}\)
\(=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)
A = 20 + 21 + 22 + 23 + 24 + 25 … + 299
A=( 20 + 21 + 22 + 23 + 24) +( 25 … + 299)
A= 20.(20 + 21 + 22 + 23 + 24)+25.( 25 … + 299)
A= 1. 31+ 25.31… + 295.31
A= 31. (1+25...+295)
KL: ......
\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)
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: \(\left[600-\left(40:2^3+3\cdot5^3\right)\right]:5\)
\(=\left[600-5-375\right]:5\)
\(=44\)
b: \(16\cdot12^2-\left(4\cdot23^2-59\cdot4\right)\)
\(=16\cdot144-4\cdot\left(23^2-59\right)\)
\(=2304-4\cdot470\)
\(=424\)
c: Ta có: \(2^{100}-\left(1+2+2^2+2^3+...+2^{99}\right)\)
\(=2^{100}-2^{100}+1\)
=1
d: Ta có: \(169\cdot2011^0-17\cdot\left(83-1702:23+1^{2012}\right)+2^7:2^4\)
\(=169-17\cdot\left(83-74+1\right)+2^3\)
\(=177-17\cdot10\)
=7
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
Trả lời :
20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 = 188
~ Học tốt ~
a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)
\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)
\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)
b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)
\(=-\frac{5}{72}:\frac{3.7}{8.9}\)
\(=-\frac{5}{72}:\frac{7}{24}\)
\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)
a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)
b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)
c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)
\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)
\(=\frac{-23}{25}+\frac{-27}{25}\)
\(=\frac{-50}{25}=-2\)
Lời giải:
a. $=-(37+63)+[25+(-25)]+(-9)=-100+0+(-9)=-(100+9)=-109$
b. $=[1+(-3)]+[5+(-7)]+....+[21+(-23)]$
$=\underbrace{(-2)+(-2)+....+(-2)}_{6}=(-2).6=-12$
c. $=-(280+20)+[-(79+21)]=-300+(-100)=-(300+100)=-400$
d. $=[-(27+43)]+[-(208+102)]=-70+(-310)=-(70+310)=-380$
e. $=(38+120)-(12+46)=158-58=100$
f. $=9+15+11+24=(9+11)+(15+24)=20+39=59$
a) 15 . (27 + 18 + 6 ) + 15 . ( 23 +12 ) = 15. 51 + 15 . 35 = 765 + 525 = 1290
b) 14 . 35 . 5 + 10 . 25 . 7 + 20 . 70 = 2450 + 1750 + 1400 = 2452450
c) 53 . ( 51 + 4 ) + 53 . ( 49 + 96 ) + 53 = 53 . 55 + 53 . 145 + 53 = 2915 . 7685 = 22401775
d) 46 . ( 27 +75 ) +8 . ( 58 + 25 ) + 8 . ( 58 + 25 ) . 2 = 46 . 102 + 8 . 83 + 8 . 83 = 4692 + 664 + 664 = 6020