It is known that the arithmetic sequence {an} satisfies A3 = 6, A4 + A6 = 20 (1) to find the general term an; (2) let {BN an} be the arithmetic sequence with the first term of 1 and the common ratio of 3, to find the general term formula of the sequence {BN} and its first n terms and TN

It is known that the arithmetic sequence {an} satisfies A3 = 6, A4 + A6 = 20 (1) to find the general term an; (2) let {BN an} be the arithmetic sequence with the first term of 1 and the common ratio of 3, to find the general term formula of the sequence {BN} and its first n terms and TN

(1) ∵ arithmetic sequence {an} satisfies A3 = 6, A4 + A6 = 20, ∵ a1 + 2D = 6A1 + 3D + A1 + 5D = 20, the solution is A1 = 2D = 2, ∵ an = 2n. (2) ∵ an = 2n, {BN an} is the arithmetic sequence with the first term of 1 and the common ratio of 3, ∵ BN − 2n = 3N − 1, ∵ BN = 3N − 1 + 2n, ∵ TN = (1 + 3 +...) +3n−1)+2(1+2+… +n)＝3n−12+n2+n．
If {an} is an arithmetic sequence with tolerance D ≠ 0, the general term is an, and {BN} is an arithmetic sequence with common ratio Q ≠ 1, we know A1 = B1 = 1, A2 = B2, A6 = B3. (1) find D and Q. (2) whether there are constants A and B, so that an = logabn + B holds for all n ∈ n *. If there is, if there is no, explain the reason
(1) Let a 2 = 1 + D = B 2 = q, a 6 = 1 + 5 d = B 3 = Q 2, q = 4 and d = 3 be obtained by solving the above two equations simultaneously. (2) suppose that there are constants A and B satisfying the equation, from an = 1 + (n-1) d = 3n-2, BN = QN-1 = 4N-1 and an = logabn + B, (3-loga4) n + loga4-b-2 = 0, ∵ n ∈ n *, ∵ 3 − loga4 = 0, loga4 − B − 2 = 0, ∵ a = 34, B = 1, so there exists
An is the arithmetic sequence whose tolerance is not zero! BN is the arithmetic sequence. B1 = A2, B2 = A5, B3 = A14, B2 + B3 + B
An means that the tolerance is not 0
B1 = A2, B2 = A5, B3 = A14
The general phase formula of B2 + B3 + B4 = 117 (1) an and BN
Let the tolerance of arithmetic sequence be D, and the tolerance of arithmetic sequence be q, then a1 + D = B1 = 3, a1 + 4D = 3q, a1 + 13D = 3Q square into, A1 = 3-D solution, q = 3 or 1 (rounding off) take q = 3 into, then d = 2, A1 = 1. Therefore, the general formula of arithmetic sequence is an = 1 + 2 (n-1) = 2N-1, and the general formula of arithmetic sequence is an = 3.3 (n-1) = 3
In the arithmetic sequence {an} with tolerance D and the arithmetic sequence {BN} with common ratio Q, A2 = B1 = 3, A5 = B2, A14 = B3, the general formula of {an} and {BN} can be obtained
Let the tolerance of arithmetic sequence be D, and the tolerance of arithmetic sequence be q, then a1 + D = B1 = 3, a1 + 4D = 3q, a1 + 13D = 3Q square into, A1 = 3-D solution, q = 3 or 1 (rounding off) into q = 3, then d = 2, A1 = 1. So, the general formula of arithmetic sequence is an = 1 + 2 (n-1) = 2N-1, and the general formula of arithmetic sequence is an = 3.3 (n-1) = 3 to the nth power!
According to the meaning of the title:
a1+d=3，
a1+4d=3q
a1+13d=3q^2
If A1 = 3, d = 0, q = 1 or A1 = 1, d = 2, q = 3, then
An = 3, BN = 3 or an = 2N-1, BN = 3 ^ n
How to calculate the acceleration and speed in the dot timer?
For example: there are three counting points in the clock, the interval between the counting points 1-2 is 4.86cm, the interval between the counting points 2-3 is 5.94cm, and the time interval is 1g30s. What is the speed of the counting point 2?
I know the acceleration is 9.72, how to calculate the speed at countable point 2?
Do you know that there is such a theorem:
In uniform acceleration, the instantaneous velocity in the middle is equal to the average velocity of the displacement
Then the velocity at 2 is the average velocity at 1-3, that is, (4.86cm + 5.94cm) / (1 / 15) = 1.62m/s
In addition, in the clock problem, the acceleration should be calculated by the difference method, otherwise the error will be relatively large
No plagiarism!
The mixed operation solution of the first rational number
(1) (-9)-(-13)+(-20)+(-2)
(2) 3+13-(-7)/6
(3) (-2)-8-14-13
(4) (-7)*(-1)/7+8
(5) (-11)*4-(-18)/18
(6) 4+(-11)-1/(-3)
(7) (-17)-6-16/(-18)
(8) 5/7+(-1)-(-8)
(9) (-1)*(-1)+15+1
(10) 3-(-5)*3/(-15)
(11) 6*(-14)-(-14)+(-13)
(12) (-15)*(-13)-(-17)-(-4)
(13) (-20)/13/(-7)+11
(14) 8+(-1)/7+(-4)
(15) (-13)-(-9)*16*(-12)
(16) (-1)+4*19+(-2)
(17) (-17)*(-9)-20+(-6)
(18) (-5)/12-(-16)*(-15)
(19) (-3)-13*(-5)*13
(20) 5+(-7)+17-10
(21) (-10)-(-16)-13*(-16)
(22) (-14)+4-19-12
(23) 5*13/14/(-10)
(24) 3*1*17/(-10)
(25) 6+(-12)+15-(-15)
(26) 15/9/13+(-7)
(27) 2/(-10)*1-(-8)
(28) 11/(-19)+(-14)-5
(29) 19-16+18/(-11)
(30) (-1)/19+(-5)+1
(31) (-5)+19/10*(-5)
(32) 11/(-17)*(-13)*12
(33) (-8)+(-10)/8*17
(34) 7-(-12)/(-1)+(-12)
(35) 12+12-19+20
(36) (-13)*(-11)*20+(-4)
(37) 17/(-2)-2*(-19)
(38) 1-12*(-16)+(-9)
(39) 13*(-14)-15/20
(40) (-15)*(-13)-6/(-9)
（41）3．28-4．76+1 ；
（42）2．75-2 -3 +1 ；
（43）42÷（-1 ）-1 ÷（-0.125）;
（44）(-48) ÷82-(-25) ÷(-6)2;
（45）- (2.5 )×(-2.4).
（46）-23 ×（-13）*2÷（- ）2；
（47）-14-（2-0.5）× ×[( )2-( )3];
（48）-1 ×[1-3×(- )2]-( - )2×(-2)3÷(- )3
（49）(0.12+0.32) ÷ [-22+(-3)2-3 ];
(50)-6.24×32+31.2×(-2)3+(-0.51) ×624.
（51） 120-36×4÷18+35
（52）10.15-10.75×0.4-5.7
（53）5.8×（3.87-0.13）
（54）+4.2×3.74 347+45×2-4160÷52
（55）32.52-（6+9.728÷3.2）×2.5 87（58+37）÷（64-9×5）
（56）[（7.1-5.6）×0.9-1.15] ÷2.5 （3.2×1.5+2.5）÷1.6
（57）5.4÷[2.6×（3.7-2.9）+0.62] 12×6÷（12-7.2）-6
（58）3.2×6+（1.5+2.5）÷1.6 （3.2×1.5+2.5）÷1.6
（59）5.8×（3.87-0.13）+4.2×3.74
（60）33.02－（148.4－90.85）÷2.5
Child, if only a few, then I don't mind helping you
But that's too much
Take your time with the computer
In fact, it's very simple. The methods are all the same
Child, you slowly calculate, pay attention to the positive and negative ha (can use the computer ah)
Press it slowly
so many! Do it yourself! You can put forward the symbol first, practice makes perfect, practice more, this is the basis, later also use, must practice!
Give examples to illustrate the relationship between the parts of addition. What about subtraction, multiplication and division
One addend = and - another addend
Subtraction = subtracted difference
Subtracted = difference + subtracted
One factor = product △ the other factor
Divisor = quotient × divisor
Divisor = divisor △ quotient
If an object moving in a straight line with constant acceleration and non-zero initial velocity arrives at point a through displacement S1 in time t, and then arrives at point B through displacement S2 in time t, the following judgment is correct ()
A. The velocity of the object at point a is S1 + s22tb. The acceleration of the object is 2s1t2c. The acceleration of the object is S2 − s1t2d. The velocity of the object at point B is 2s2 − s1t
A. Because the average velocity in a period of time is equal to the instantaneous velocity at the middle time, then the velocity at point a VA = S1 + s22t. So a is correct. B. according to S2 − S1 = at2, a = S2 − s1t2. So B is wrong, C is correct. D. the velocity of the object at point B VB = VA + at = S1 + s22t + S2 − s1t = 3s2 − s12t. So D is wrong
How to divide a positive number by a negative number?
Such as the title
For example, 3 / (- 3) =? 10 / (- 5) =? And so on
The result is negative, and then it is divided as positive, and the result is put after the negative sign
a>=0,b>0:-(a/b)=(-a)/b=a/(-b)=-|a/b|=-|a|/|b|
Who can help me to make two decimal points?
0.37×2.9
0.56×0.08
Can upload pictures!
0.37 0.56
* 2.9 *0.08
—— ——
333 0.0448
Seventy-four
——
one point zero seven three