In the known sequence {an}, A1 = 1, A2 = 2, and an + 1 = (1 + Q) an-qan-1 (n ≥ 2, Q ≠ 0). An + 1, an-1 are lower subscripts (1) Let BN = an + 1-an (n ∈ positive integer), it is proved that {BN} is the general formula for finding {an} in the equal ratio sequence (2). An + 1 and an-1 are the subscripts

In the known sequence {an}, A1 = 1, A2 = 2, and an + 1 = (1 + Q) an-qan-1 (n ≥ 2, Q ≠ 0). An + 1, an-1 are lower subscripts (1) Let BN = an + 1-an (n ∈ positive integer), it is proved that {BN} is the general formula for finding {an} in the equal ratio sequence (2). An + 1 and an-1 are the subscripts

A (n + 1) = (1 + Q) an QA (n-1) a (n + 1) = an + QaN QA (n-1) a (n + 1) - an = QaN QA (n-1) [a (n + 1) - an] / [an-a (n-1)] = q because BN = a (n + 1) - an, so BN / b (n-1) = q, so BN is an equal ratio sequence with Q as the common ratio, BN = (a2-a1) * q ^ (n-1) BN = q ^ (n-1) a (n + 1) - an = q ^ (n-1) a (n + 1) -

In the sequence {an}, it is known that A1 = - 2, a (an-1) subscript = 2An + N + 1 ((an-1) and N are subscripts)
(1) Proof: the sequence {a (n + 1) - an + 1} is equal ratio sequence
(2) Find the general term formula of sequence {an}
(3) Sum: SN = | A1 | + | A2 | + | A3 | +... | an|

I can't even write the title clearly, so how to answer it

Let (1 + x) + (1 + x) ^ 2 + (1 + x) ^ 3 + +(1 + x) ^ n = A0 + a1x +... A (n-1) x ^ (n-1) + anx ^ n, if a (n-1) = 2011, then A0 + A1 + A2 + +A (n-1) + a (n) is equal to?
A (2^2010)-2 B (2^2011)-2
c (2^2012)-2 C (2^2011)-1

Suppose x = 1, then (1 + x) + (1 + x) ^ 2 + (1 + x) ^ 3 + +(1+x)^n=2^1+2^2+2^3+…… +2 ^ n is an equal ratio sequence
The formula can be expressed as Sn = A1 (1-Q ^ n) / (1-Q) A1 = 1 + 1 = 2, q = 2, and a (n-1) = 2011 is the 2012 term
n=2012
(2^2012)-2

It is known that s (x) = a1x + a2x ^ 2 +... + anx ^ n, and A1, A2,..., an form the arithmetic sequence, and N is a positive even number
Let s (1) = n ^ 2, s (- 1) = n
(1) Finding the general term formula of sequence {an}
(2) Proof s (1 / 2)

（1）
Because A1, A2,..., an constitute the arithmetic sequence, so let the tolerance be d,
And because s (1) = n ^ 2, then
A1 + A2 +... + an = n ^ 2, then it can be obtained from the summation formula Sn = [2Na1 + n (n-1) D] / 2
[2Na1 + n (n-1) D] / 2 = n ^ 2
A1 + D (n-1) / 2 = n, that is, A1 = n (1-D / 2) + D / 2,
Because A1 is the first term of arithmetic sequence, it has nothing to do with N, so 1-D / 2 = 0, that is, d = 2
So A1 = 1, we get an = a1 + (n-1) d, and bring in an = 1 + (n-1) * 2,
The general formula of {an} is an = 2N-1;
（2）
From (1), s (x) = x + 3x ^ 2 + 5x ^ 3 +... + (2n-1) x ^ n, then
S(1/2)=(1/2)+3*(1/2)^2+5*(1/2)^3+...+(2n-1)*(1/2)^n,
(1 / 2) s (1 / 2) = (1 / 2) ^ 2 + 3 * (1 / 2) ^ 3 + 5 * (1 / 2) ^ 4 +... + (2n-1) * (1 / 2) ^ (n + 1), then
S (1 / 2) - (1 / 2) s (1 / 2) = (1 / 2) + 2 * (1 / 2) ^ 2 + 2 * (1 / 2) ^ 3 +... + 2 * (1 / 2) ^ n - (2n-1) * (1 / 2) ^ (n + 1)
(1/2)S(1/2)=-(1/2)+(1-(1/2)^n)/(1-(1/2))-(2n-1)*(1/2)^(n+1),
S(1/2)=-1+4-(1/2)^(n-2)-(2n-1)*(1/2)^n,
S(1/2)=3-(1/2)^(n-2)-(2n-1)*(1/2)^n,
So s (1 / 2)

Known (1 + x) ^ n = A0 + a1x + a2x ^ 2 + +anx^n,
a1+2a2+3a3+… +Nan = 80. Find n

The derivation of the two sides of the equation, the equation and the derivative of a1 + 2A2 + 3a3 + +Nan (x = 1), n (1 + x) ^ (n-1) on the left
N (1 + x) ^ (n-1) = 80 (when x = 1), n * 2 ^ (n-1) = 80, n = 5

(x-1) ^ n = A0 + a1x ^ 1 + a2x ^ 2 + a3x ^ 3 +... + anx ^ n, find A0 + A1 + A2 +... + an =?

（x-1）^n=a0+a1x^1+a2x^2+a3x^3+...+anx^n,
When x = 1, (1-1) ^ n = A0 + A1 + A2 +.. + an = 0;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
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Let f (x) = (2x-1) to the third power, and f (x) expand to the cubic form of = A0 + a1x = the square of a2x + ax, try to find A0 + A1 + Q2 + a3; a0-a1 + A2-A3

Replace x = 1 and x = - 1 to the original equation, f (1) = A0 + A1 + A2 + a3 = 1, f (- 1) = a0-a1 + A2-A3 = - 27

It is known that (1 + x) + (1 + x) 2 + (1 + x) 3 + +（1+x）n=a0+a1x+a2x2+a3x3+… +Anxn, if a1 + A2 + +An-1 = 29-n, then n is a positive integer=______ ．

Only in the expansion of (1 + x) n can there be a term containing xn, its coefficient is 1, let x = 0 get A0 = n, let x = 1 get A0 + A1 + A2 + + an-1 + an = 2 + 22 + 23 + + 2n = 2n + 1-2, ∧ A1 + A2 + + an-1 = 2n + 1-2-1-n ∧ 2n + 1-3-n = 29-n get n = 4; so the answer is 4

In the known sequence {an}, A0 = A1 = 1, and the root sign ana (n-2) - root sign a (n-1) a (n-2) = 2A (n-1), we can find the general term formula of the sequence {an}
N-1 and n-2 in brackets are corner marks

If √ ana (n-2) - √ a (n-1) a (n-2) = 2A (n-1) (n ≥ 2), both sides of the original formula are divided by a (n-1) at the same time to get √ [ana (n-2) / a (n-1) ^ 2] - √ [a (n-2) / a (n-1)] = 2, let BN = √ [an / a (n-1)], then B1 = √ (A1 / A0) = 1, so BN / b (n-1) - 1 / b (n-1) = 2, that is BN = 2B (n-1) + 1 (n ≥ 2), so BN + 1 = 2 [b (n-1)]

Given the sequence an satisfies A1 = 1 / 2, ana (n-1) = a (n-1) - an to find the general term formula of an
To complete the problem-solving process, the best text
On the left is an times a (n-1)

Two sides divide ana (n-1)
Available
1=1/an-1/a(n-1)
So {1 / an} is an arithmetic sequence, and the tolerance D is 1
The first term is 1 / A1 = 2
So 1 / an = a1 + (n-1) d = 2 + (n-1) = n + 1
an=1/(n+1)