It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!

It is known that the equal ratio sequence satisfies A1 = 2, A2 + a3 = 12 and A4 > 0.1. Find the general term formula of sequence an. 2. If BN = anlog2an, the first n term of sequence BN is SN The main thing is to ask the second question!

1, let the common ratio be q, then A2 = A1 * q = 2q, A3 = A1 * Q & # 178; = 2q & # 178;
So A2 + a3 = 2q + 2q & # 178; = 12, so Q & # 178; + q-6 = 0, (Q-2) (Q + 3) = 0
And A4 = A1 * Q & # 179; = 2q & # 179; > 0, so Q > 0, so Q-2 = 0, q = 2
Then an = A1 * q ^ (n-1) = 2 × 2 ^ (n-1) = 2 ^ n (n ∈ n +)
2,bn=2^n×log2(2^n)=n×2^n
So Sn = 1 × 2 ^ 1 + 2 × 2 ^ 2 + 3 × 2 ^ 3 + +n×2^n ①
So 2Sn = 1 × 2 ^ 2 + 2 × 2 ^ 3 + +(n-1)×2^n+n×2^(n+1) ②
① - 2, get: - Sn = 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + +2^n-n×2^(n+1)
=2(1-2^n)/(1-2)-n×2^(n+1)
=2^(n+1)-2-n×2^(n+1)
=(1-n)×2^(n+1)-2
So Sn = (n-1) × 2 ^ (n + 1) + 2
What does the sign diag mean in linear algebra
A diagonal matrix, such as diag (1,2,3), denotes a diagonal matrix with diagonal elements of 1,2,3
Diag is (extract diagonal elements)
There are also linear algebraic functions
Det, inv, QR, SVD, bdiag, spec, Schur, trace
In △ ABC, ∠ B = 90 °, ab = 7, BC = 24, AC = 25
A. 2B. 3C. 4D. 5
Let PE = pf = PG = XS △ ABC = 12 × ab × CB = 84, s △ ABC = 12ab × x + 12ac × x + 12bc × x = 12 (AB + BC + AC) · x = 12 × 56x = 28x, then 28x = 84, x = 3
The problem of discontinuities: y = x ^ 3-x / sin π X; y = x ^ 2-x / SiNx (x ^ 3-1); y = x ^ 2-x / | SiNx | (x ^ 2-1) how many discontinuities of the first kind
1,1.1
It is known that the sequence {an} is the arithmetic sequence, a1 + A2 + a3 = 15, and the sequence {BN} is the proportional sequence, b1b2b3 = 27
If A1 = B2, A4=
B3, find the formula of sequence BN and an
Equal ratio sequence, b1b2b3 = 27
Then B2 = 3
Then A1 = 3
Arithmetic sequence, a1 + A2 + a3 = 15,
Then A2 = 5
So the tolerance of arithmetic sequence {an} is 2
General term formula an = 1 + 2n
a4 = 9 = b3
So the equal ratio sequence {BN}, the common ratio is 3
General term formula BN = 3 ^ (n-1)
What does alpha mean with two absolute symbols? Linear algebra
The norm of V is a function p: V → R; X → P (x), which satisfies the following conditions: a ∈ F, & ᦉ 8704; u, V ∈ V, 1. P (V) ≥ 0
In RT triangle ABC, angle c = 90 degrees, BC: AC = 8:15, find three triangle functions of angle A
Let BC = 8a and AC = 15A
Then AB = √ [(8a) ^ 2 + (15a) ^ 2]
=17a
Then Sina = 8A / 17a = 8 / 17
cosA=15a/17a=15/17
tanA=8a/15a=8/15
cotA=15a/8a=15/8
Find the discontinuous point of F (x) = (SiNx + x) / SiNx, and prove the type of discontinuous point!
Curriculum reform is too far off the mark, after a long time have forgotten, should be seeking 1 / SiNx discontinuity
I don't know how to write it. This is an example
y=1/sinx
Domain: SiNx ≠ 0
Equivalent to: X ≠ 0, and SiNx ≠ 0, that is, X ≠ K π (k is an integer)
In other words, when x = k π (k is an integer), the function is undefined
And when x tends to K π (k is an integer), the function value tends to inf
So x = k π (k is an integer) is the infinite discontinuity of a function
Given that the sequence {an} is an arithmetic sequence, and A1 = 2, a1 + A2 + a3 = 12, let BN = 3 ^ an, prove that the sequence {BN} is an equal ratio sequence
Let the tolerance value be c
a1+a2+a3=a1+（a1+c）+(a1+c+c)=3a1+3c=12
C=2
an=a1+c(n-1)=2n
bn=3^(2n)
b(n+1)/bn=3^(2n+2)/3^2n=9
So BN is an equal ratio sequence
12=3a2==>a2=4
d=a2-a1=2
an=2n
bn=3^(2n)
bn-1=3^(2(n-1))
bn/b(n-1)=3^[2n-2(n-1)]=3^2=9
So the sequence {BN} is an equal ratio sequence;
Find the general solution of the equations: ① X1 + x2 + 2x3-3x4 = 0; ② 3x1 + 2x2-x3 + 2x4 = 0; ③ x1-5x3 + 8x4 = 0
x1 = 5 x3 - 8 x4
x2 = -7 x3 + 11 x4
X3x4 is a free variable, which can be taken from R at will