If limf (x0-2 △ x) - f (x0 + 3 △ x) / Δ x = 1, then f '(x0)=

If limf (x0-2 △ x) - f (x0 + 3 △ x) / Δ x = 1, then f '(x0)=

∵limf(x0-2△x)-f(x0+3△x)/△x=1∴limf(x0-2△x)-f(x0）+f(x0)-f(x0+3△x)/△x=1∴lim[f(x0-2△x)-f(x0)]/Δx-lim[f(x0+3△x)-f(x0)/△x=1∴(-2)lim[f(x0-2△x)-f(x0)]/(-2Δx)-3lim[f(x0+3△x)-f(x0)/(3△x)=1∴-...

If f (x) = 1 / x + 1, then f (x + 1) is equal to

f(x+1)=1/(x+1)+1 =1/x+2

Let f (log2 base x) = 2 ^ x (x is greater than 0), then f (3) is equal to

Let log2 x = t, then x = 2 ^ t
So f (T) = 2 ^ (2 ^ t)
So f (3) = 2 ^ (2 ^ 3) = 2 ^ 8 = 256

It is known that {an} is an increasing sequence of equal proportion numbers, and {A1, A3, A5} belongs to {- 10, - 6, - 2,0,1,3,4,16}, (1) the general term formula of {an}, (2) whether there is an arithmetic sequence {BN}, such that a1bn + A2B (n-1) + a3b (n-2) + +Anb1 = 2 ^ (n + 1) - n-2 is true for all integers n belongs to? If it exists, find B. if it does not exist, explain the reason

(1) It is known that {an} is an increasing sequence of equal proportion numbers. It can be seen that equal proportion cannot be negative. There are two cases
If Q

If the common ratio of the sequence {an} is - 1 / 2, then LIM (a1 + A2 +... + an) / (A2 + A4 +...)+

=lim(a1+a2+...+an)/[(a1+a2+...+an)·(-1/2)]
=-2

Given that the sequence A0, A1, A2,..., an,..., satisfies the relation (3-A (n + 1)) (6 + an) = 18, and A0 = 3, what is the value of 1 / A1 +. + 1 / AI
Note: a (n + 1) is a number

From [3-A (n + 1)] (an + 6) = 18 → a (n + 1) = 3an / (an + 6) -- (1) adding 3 on both sides, we get a (n + 3) + 3 = 6 (an + 3) / (an + 6) -- (2) / ① → [a (n + 3) + 3] / a (n + 1) = 2 (an + 3) / an, so the sequence {(an + 3) / an} is an equal ratio sequence with the first term of 2 and the common ratio of 2

If the sum of the first n terms of the arithmetic sequence {an} is Sn, then LIM (n tends to infinity) [Sn / (an ^ 2)] =?

a(n)=a+(n-1)d,[a(n)]^2=[a+(n-1)d]^2=[nd+a-d]^2,
s(n)=na+n(n-1)d/2=(d/2)n^2+n(a-d/2),
When d = 0, a (n) = a, s (n) = Na
When d = 0, a = 0, s (n) / [a (n)] ^ 2 has no meaning
When d = 0 and a is not equal to 0, s (n) / [a (n)] ^ 2 = n / A, LIM (n - > infinity) {s (n) / [a (n)] ^ 2} = (n - > infinity) {n / a} = positive infinity
When D is not equal to 0,
s(n)/[a(n)]^2=[(d/2)n^2+n(a-d/2)]/[nd+a-d]^2=[d/2+(a-d/2)/n]/[d+(a-d)/n]^2,
LIM (n - > infinity) {s (n) / [a (n)] ^ 2} = LIM (n - > infinity) {[D / 2 + (A-D / 2) / N] / [D + (A-D) / N] ^ 2} = (D / 2) / D ^ 2 = 1 / (2D)

If the sequence {an} satisfies A1 = 1, an + 1 · an = 2 ^ n, then s2012=

Consider n + 1: an + 2An + 1 = 2 ^ (n + 1) to know an + 2 / an = 2
Even numbers are equal to odd numbers After that, I don't know the simple formula

If the sequence an satisfies an + 1 / an = n + 2 / N and A1 = 1, then an=

∵a(n+1)/an=（n+2）/n,a1=1,
∴an=(n+1)/(n-1)*a(n-1)
=(n+1)n/[(n-1)(n-2)]*a(n-2)
.
=(n+1)n(n-1).4*3/[(n-1)(n-2).2*1]*a1
=n(n+1)/2

Given that the sequence {an} satisfies A1 = 1, a (n + 1) an = 2 ^ n, Sn is the sum of the first n terms of the sequence {an}, what is s2012?

a2a1=2 a2=2/a1=2/1=2
a(n+1)an=2ⁿ a(n+2)a(n+1)=2^(n+1)
[a (n + 2) a (n + 1)] / [a (n + 1) an] = a (n + 2) / an = 2 ^ (n + 1) / 2 & # 8319; = 2, is a fixed value. The odd term of a sequence is an equal ratio sequence with 1 as the first term and 2 as the common ratio, and the even term is an equal ratio sequence with 2 as the first term and 2 as the common ratio
S2012=(a1+a3+...+a2011)+(a2+a4+...+a2012)
=1×(2^1006 -1)/(2-1) +2×(2^1006 -1)/(2-1)
=3×2^1006 -3