In {an}, the first n terms of A1 = 2 and Sn satisfy Sn + 1 + sn-1 = 2Sn + 1 (1) to find the general term formula of sequence an (2) the n power of BN = 4 + the N-1 power of negative 1 multiplied by the an power of 2

In {an}, the first n terms of A1 = 2 and Sn satisfy Sn + 1 + sn-1 = 2Sn + 1 (1) to find the general term formula of sequence an (2) the n power of BN = 4 + the N-1 power of negative 1 multiplied by the an power of 2

Sn+1+Sn-1=2Sn+1 (Sn+1-Sn)+(Sn-1-Sn)=1 (an+1)-an=1
So arithmetic sequence can be done next, right
Sn+1+Sn-1=2Sn+1 (Sn+1-Sn)+(Sn-1-Sn)=1 (an+1)-an=1
an=n+1
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If A1, A2, A3 and A4 are known to be linearly independent, then ()
(A) A1 + A2, A2 + a3, A3 + A4, A4 + A1 are linearly independent
(B) A1-a2, A2-A3, a3-a4, a4-a1 are linearly independent
(C) A1 + A2, A2 + a3, A3 + A4, a4-a1 are linearly independent
(D) A1 + A2, A2 + a3, a3-a4, a4-a1 are linearly independent
(C) Right
Because of the determinant
1 1 0 0
0 1 1 0
0 0 1 1
1 0 0 -1
=- 2 (not equal to 0)
Let f (x) = (x square + ax + b) e x power (x ∈ R), if a = 2, B = - 2, find the extremum of F (x)
F (x) = (x ^ 2 + 2x-2) e ^ X; derivation: F '(x) = (2x + 2) e ^ x + (x ^ 2 + 2x-2) e ^ x = (x ^ 2 + 4x) e ^ X; Let f' (x) = 0; the solution is x = 0 or - 4; similarly, f '' (x) = (2x + 4) e ^ x + (x ^ 2 + 4x) e ^ x = (x ^ 2 + 6x + 4) e ^ X; f '' (0) = 4 > 0; F '' (- 4) = - 4 * e ^ (- 4)
How can Casio fx-4500PA calculator convert 12.58244 degrees into degrees, minutes and seconds
1）37°20’+54.8°= 92°8'
2）70°42’45”+82°37’25”= 153°20'10”
3）120°-36°18’52”=83°41’8”
4）82.7°= 82°42’
5）20°18’= 20.3°
6）57°30’-27.5°=30°
7）48°19’+67°21’= 115°40’
8）27°40’+15.3°= 42°58’
9）100°15’-79°26’= 20°49’
10)80°24’/3=26°48’
（11）180°-46′42〃＝179°60′-46′42〃＝179°59′60〃-46′42〃＝179°13′18〃
（12）28°36′+72°06′＝100°42′
（13）50°24′×3＝150°72′＝151°12′
（14）49°28′52〃÷4＝12.25°07′13〃＝12°（0.25×60+07）′13〃＝12°22′13〃
(15) 17 degrees 25 minutes 18 seconds * 3 = 51 degrees + 75 minutes + 54 seconds = 52 degrees 15 minutes 54 seconds (60 minutes into 1 degree)
90 degrees - 52 degrees 15 minutes 54 seconds = 89 degrees + 59 minutes + 60 seconds - 52 degrees - 15 minutes - 54 seconds = 37 degrees + 44 minutes + 6 seconds = 37 degrees 44 minutes 6 seconds
(16) 40 degrees 27 points / 3 = 13 degrees + (1 degree + 27 points) / 3 = 13 degrees + 87 points / 3 = 13 degrees 29 points
（17）180°- 46′42〃＝179°60′- 46′42〃＝179°59′60〃-46′42〃＝179°13′18〃
（18）28°36′+7 2°06′＝100°42′
（19）50°24′×3＝150°72′＝151°12′
（20）49°28′52〃÷4
＝12.25°07′13〃
＝12°（0.25×60+07）′13〃
＝12°22′13〃
52°45′-32°46′=19°59′
79°16′30〃-7°25′+12°38′30〃=66°38′0〃
1）51°37′-32°45′31〃
=51°36′60‘’-32°45′31〃
=18°51‘29’‘
（2）35°35'35''×5
=175°175’175‘’
=177°57‘55’‘
（3）（180°-91°32'24''）÷2
=88°27’36‘’/2
=4°23‘48’‘
（4）176°51'÷3
=58°57’
124 degrees 16 minutes 12 seconds = () degrees
1 second = 1 / 60 minutes
So 12 seconds = 12 / 60 = 0.2 minutes
So 124 degrees 16 minutes 12 seconds = 124 degrees 16.2 minutes
1 point = 1 / 60 degree
So 16.2 points = 16.2 / 60 = 0.27 degrees
So 124 degrees 16 minutes 12 seconds = 124 degrees 16.2 minutes = 124.27 degrees
48°39'+67°41'=
48°39'+67°41'=48°+67°+（39/60+41/60）°=116°20'
-78°19'40''=
-78°19'40''=-{78°+（19/60）°+（40/3600）°}=78.328°
51°37′-32°45′31〃
=51°36′60‘’-32°45′31〃
=18°51‘29’‘
35°35'35''×5
=175°175’175‘’
=177°57‘55’‘
（180°-91°32'24''）÷2
=88°27’36‘’/2
=4°23‘48’‘
176°51'÷3
=58°57’
72°35'/2
=36°17.5'=36°17'30''
18°33'*4
=72°132'=72°+2°12'=74°12'
57.18°
=57°10'48''
Degrees, minutes and seconds are carried by 60. Minutes divided by 60 are converted into degrees. Seconds divided by 60 are converted into minutes. Seconds are directly converted into degrees divided by 60 * 60. If you master this concept, you will be able to solve such type problems
It is known that the sum of the first three terms of the set of arithmetic sequence an is 6, and the sum of the first eight terms is - 4. Find the general term formula of the set of arithmetic sequence an? Let BN = (4-an) * q to the power of n-1, find the sequence
The first n terms and Sn of BN set
A 1 = 3 and d = - 1 are obtained from an series of equations
an=4-n
bn=nq^(n-1)
sn=1+2q+3q^2+4q^3+.+nq^(n-1)
kn=n+(n-1)q+(n-2)q^2+(n-3)q^3+.+2q^(n-2)+q^(n-1)
=(1-q^n)/(1-q)+(1-q^(n-1))/(1-q)+.+(1-q^2)/(1-q)+(1-q)/(1-q)
=(n-(q+q^2+.+q^n))/(1-q)
=(n(1-q)-q+q^(n+1))/(1-q)^2
sn+kn=(n+1)(1-q^n)/(1-q)
sn=(n+1)(1-q^n)/(1-q)-kn=(1-nq^n-q^n+nq^(n+1))/(1-q)^2
Let the vector groups A1, A2 and A3 be linearly related, and the vector groups A2, A3 and A4 be linearly independent. It is proved that A1 can be linearly represented by A2 and A3
It shows that the vector group A1, A2, A3, A4 are linearly related; that is, there are four numbers K1, K2, K3, K4 that are not all zero, such that K1 * a1 + K2 * A2 + K3 * A3 + K4 * A4 = 0 (Note: because it is not easy to write subscript here, it is declared that K1, K2, K3, K4 are coefficients) and because A4 cannot be expressed linearly by A1, A2, A3, there is no following equation: A4 = C
If the function f (x) = 13x3-x2 + AX-1 has an extreme point, then the value range of a is ()
A. （-∞，0）B. （-∞，0]C. （-∞，1）D. （-∞，1]
The function f (x) = 13x3-x2 + AX-1 has extremum, the derivative f ′ (x) = x2-2x + a = 0 of F (x) has two real roots, and the derivative f ′ (x) = 4-4a ＞ 0, a ＜ 1
Can the sine and cosine of some angles not be calculated with a pen, but only with a calculator?
Yes, the general exam only tests some special values
Most of them use computers
I can't figure it out. If I can't use the computer in the exam, I will tell you
Calculated with the knowledge of advanced mathematics,
That's how the math watch is made
But the process is extremely complicated
Let the sequence {an} satisfy A1 = 2, a (n + 1) - an = 3 times the (2n-1} power of 2. 1. Find the general term formula of the sequence; 2. Let BN = n times an, find the sum of the first n terms of the sequence
I'm not good at math. I have trouble writing the process
This is a common sequence problem, so it's easy to understand it once. 1) as follows: a (n + 1) - an = 3 * 2 ^ (2n-1) an-a (n-1) = 3 * 2 ^ (2 (n-1) - 1. A3-a2 = 3 * 2 ^ (2 * 2-1) a2-a1 = 3 * 2 ^ (2 * 1-1) add all to get: a (n + 1) - A1 = 3 * [2 ^ (2n-1) + 2 ^ (2 (n-1) - 1 +. + 2 ^ (2 * 2-1) + 2 ^ (2 * 1-1)]
Give any vector A1, A2, A3, A4, and prove that a1 + A2, A2 + a3, A3 + A4, A4 + A1 are linearly related. I hope there is a complete solution
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