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S1 = \(1+2+2^2+2^3+...+2^{62}+2^{63}\)
2 . S1 = \(2+2^2+2^3+2^4+...+2^{63}+2^{64}\)
2.S1 - S1 =\(\left(2+2^2+2^3+2^4+...+2^{63}+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{62}+2^{63}\right)\)
S1 = \(2^{64}-1\)
\(S=1+2+2^2+2^3+...+2^{62}+2^{63}\)
\(2S=2+2^2+2^3+2^4+...+2^{63}+2^{64}\)
\(2S-S=2^{64}-1\)
\(S=2^{64}-1\)
A = 1 + 31 + 32 + 33 + ... + 320
3A = 3( 1 + 31 + 32 + 33 + ... + 320 )
3A = 3 + 32 + 33 + 34 + ... + 321
3A - A = ( 3 + 32 + 33 + 34 + ... + 321 ) - ( 1 + 31 + 32 + 33 + ... + 320 )
=> 2A = 3 + 32 + 33 + 34 + ... + 321 - 1 - 31 - 32 - 33 + ... - 320
2A = 2 + 321
A = \(\frac{2+3^{21}}{2}\); B = \(\frac{3^{21}}{2}\)
Vì 2 + 321 > 321
=> \(\frac{2+3^{21}}{2}\)> \(\frac{3^{21}}{2}\)hay A > B
A=1+ 31+32+33+...+320
3A = 3 + 3^2 + 3^3 + ... + 3^21
2A = 3^21 - 1
A = 3^21 - 1/2
3^21-1 < 3^21
=> 3^21-1/2 < 3^21/2
=> A < B
\(S=1+2+2^2+2^3+...+2^{10}\)
\(S=1+2\left(2+2^2+...+2^9\right)\)
\(S=1+2\left(S-2^{10}\right)\)
\(S=1+2S-2^{11}\)
\(S=2^{11}-1\)
2S= 2+22+....+211
2S-S=(2+22+....+211)-(1+2+....+210)
S=211 - 1
dễm
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(=3^0-3^1+3^2-3^3+...+3^{98}-3^{99}\)có 100 hạng tử
\(=\left(3^0-3^1+3^2-3^3\right)+\left(3^4-3^5+3^6-3^7\right)+...+\left(3^{96}-3^{97}+3^{98}-3^{100}\right)\) có 25 cặp
\(=-20+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)\)
\(=-20\left(1+3^4+...+3^{96}\right)⋮-20\)
Ta có :
27 mũ 11 = (3mu3)mũ11=3 mũ33
81 mũ 8 = (3 mũ 4)mũ 8 =3 mũ 32
Vì 3 mũ 33 >3 mũ 32
Vậy 27 mũ 11 > 81 mũ 8
Cho xin k
HOK TỐT
Phần a sai đề nha
b) S = 3 + 32 + 33 + 34 + ............ + 320
S = ( 3 + 32 ) + ( 33 + 34 ) + ........... + ( 319 + 320 )
S = 3 . ( 1 + 3 ) + 33 . ( 1 + 3 ) + ....... + 319 . ( 1 + 3 )
S = 3 . 4 + 33 . 4 + ............. + 319 . 4
S = 12 + 27 . 4 + ........... + 319 . 4
S = 12 + 108 + ........... + 319 . 4
Mà 12 ; 108 \(⋮\) 12 \(\Rightarrow\) ( 12 + 108 + ............ + 319 . 4 ) \(⋮\) 12
Vậy S \(⋮\) 12 ( ĐPCM )
b/S=3+3^2+3^3+3^4+......+3^20(gồm 21 số hạng)
S=(3+3^2)+(3^3+3^4)+(3^5+3^6)+......+(3^19+3^20)
S=1(3+3^2)+3^2(3+3^2)+......+3^18(3+3^2)
S=1.12 +3^2.12 +........+3^18.12
S=12.(1+3^2+3^4+......+3^18)
Vậy S chia hết cho 12
Ta có : C = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 310 + 311
= (1 + 3 + 32 + 33) + (34 + 35 + 36 + 37) + (38 + 39 + 310 + 311)
= (1 + 3 + 32 + 33) + 34.(1 + 3 + 32 + 33) + 38.(1 + 3 + 32 + 33)
= 40 + 34.40 + 38.40
= 40.(1 + 3 + 32 + 33)
= 10.4.(1 + 3 + 32 + 33) \(⋮\)10
=> \(C⋮10\left(\text{ĐPCM}\right)\)
A=1+32+34+.....+32008
32A-A=1+32+34+....+32008+32010-[1+32+34+...+32008]
9A-A=32010
8A=32010
Mình làm vậy đúng hay sai.
A=1+32+33+34+......+32008
3A=3+33+34+35+......+32009
3A-A=(3+33+34+35.....+32009)-(1+32+33+34+...+32008)
A=(3+32009)-(1+32008)=(3+31+32008)-(1+32008)=3+3-1=5
bạn ơi qua giúp mk vs
ta có:
\(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}\)
\(3S=3\times\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}\right)\)
\(3S=\frac{3}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{3}{3^{20}}\)
\(3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}\)
\(3S-S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}\right)\)
\(2S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-....-\frac{1}{3^{20}}\)
\(2S=1-\frac{1}{3^{20}}\)
\(S=\frac{1-\frac{1}{3^{20}}}{2}\)
\(S=0,499999999999999.....\)(Mik bấm máy tính thấy vậy)