In the sequence {an}, a1 + A6 = 33, a3a4 = 32, a (n + 1)

In the sequence {an}, a1 + A6 = 33, a3a4 = 32, a (n + 1)

1 because it is an equal ratio sequence, a3a4 = a1a6 = 32
So A1, A6 are x ^ - 33x + 32 = 0.2 (Veda)
x1=32 x2=1
a(n+1)

Let BN = log2a1 + log2a2 + log2a3 +... + log2an
(1) Finding the first n terms of {1 / BN} and TN
(2) If 7 + Kan ≥ (n + 1), TN is constant for any positive integer n, find the range of K
PS: the logarithm of log with the base of 2 in the title above

(1)
∵ {an} is an equal ratio sequence, let the common ratio be Q
a1=2,4（a3）²=a2*a6
∴4(2q²)²=2q*2q^5
∴q²=4
∵an>0
∴q>0,q=2
∴an=2^n
bn=log₂a1+log₂a2+log₂a3+...+log₂an
=log₂(a1*a2*a3*.*an)
=log₂2^(1+2+3+...+n)
=1+2+3+.+n
=(n+1)n/2
∴1/bn=2/[n(n+1)]=2[1/n-1/(n+1)]
The sum of the first n terms of {1 / BN}
Tn=2[1-1/2+1/2-1/3+.+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
(2)
7 + Kan ≥ (n + 1), TN holds for any positive integer n
I didn't make it clear
7 + Kan ≥ (n + 1) TN?
7+k*2^n≥(n+1)*2n/(n+1)=2n
∴k*2^n≥2n-7
That is, K ≥ (2n-7) / 2 ^ n is constant
It needs K ≥ [(2n-7) / 2 ^ n] max
Note CN = (2n-7) / 2 ^ n
c1,c2,c3>0
When n ≥ 4, CN > 0
From C (n + 1) / cn ≥ 1
That is {[2 (n + 1) - 7] / 2 ^ (n + 1)} / [2n-7) / 2 ^ n] ≥ 1
(2n-5) / (4n-14) ≥ 1
2n-5≥4n-14
2n≤9
n≤9/2
∴n≤4
∴c1

If we know the first term A1 = 1 and the common ratio q = 2, then log2a1 + log2a2 + +log2a11=（　　）
A. 46B. 35C. 55D. 50

The first term of {an} is A1 = 1, the common ratio is q = 2,... Log2a1 + log2a2 + +log2a11=log2（a1a2… a11）=log2(a110q1+2+3+… +10) So the answer is: 55

If the sequence {an}, A4 · A7 = 8, log2a1 + log2a2 + +log2a10=（　　）
A. 5B. 10C. 15D. 20

By the equal ratio sequence {an}, A4 · A7 = 8, ｜ A10 = a2a9 = =a4a7=… =8．∴log2a1+log2a2+… +log2a10=log2（a1a2•… So the answer is: 15

In the positive proportional sequence, a4a5 = 32, log2a1 + log2a2 +... + log2a8 =?

log2a1+log2a2+...+log2a8=log2[a1*a2*..*a8]=log2(32^4)=4log2(32)
=4log2(2^5)=20

Equal ratio sequence TN = log2a1 + log2a2 + +log2an
Find the value of T7
How did the A4 downstairs work out to the seventh power

The seventh power of log2a4
TN=log2a1+log2a2+… +log2an
T7=log2a1+log2a2+… +log2a7=log2a1a2… a7
So T7 = log2a4 to the seventh power

It is known that the equal ratio sequence {an} satisfies an ＞ 0, and A5 * a2n-5 = 2 ^ 2n (n > = 3 and N belongs to n *), then when n ＞ = 1, log2 (A1) + log2 (A3) + +log2（a2n-1）=?
What methods should be used and what is the breakthrough?
Thank you very much=^=

∵ a [n] is an equal ratio sequence. There is also a [5] * a [2n-5] = 2 ^ 2n ∵ a [1] * a [2N-1] = 2 ^ 2n. Let a [n] have the common ratio Q. ∵ a [1] * q ^ n-1 = a [n], a [2N-1] = a [1] * q ^ 2n-2, ∵ a [1] * a [2N-1] = (a [1] * Q * n-1) & ∵ 178; ∵ a [n] & ∵ 178; = 2 ^ 2n, thus a [n] = 2 ^ n