Let the sum of the first n terms of the sequence an be Sn, and Sn = n ^ 2-4n + 4, BN = an / 2 ^ n. can we use dislocation subtraction to find the sum of the first n terms of BN and TN?

We can use dislocation subtraction. First, we use SN to find an, but the expression of an is divided into n = 1 and N is not equal to 1, and then substitute it into BN. When we use dislocation subtraction, we can start from n = 2 to N, and then add in the case of n = 1. My result is: TN = 1 - (2n-1) / 2 ^ n

High school mathematics problems about equal ratio sequence
Proportional sequence: if A9 + A10 = a, A10 + A20 = B (a, B is not equal to zero), find the value of A99 + A100
It's a process

a10+a20=(a9+a10)d=b=ad
Then d = B / A
Then A99 + A100 = (A9 + A10) * 10d = a * 10d = 10ad

High school mathematics problems -- equal ratio series
1. Given that the equal ratio sequence {an} satisfies a1 + A2 = 3, A2 + a3 = 6, then A7 =?
2. In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and 2b = a + C, B = 30 ° and the area of △ ABC is 3 / 2. What is B equal to?
Detailed process required, thank you!

1、 If we divide a form into two, we get q = 2. If we substitute a form into two, we get A1 = 1, a7 = 64
2、 If we substitute 2B = a + C into the formula, we can get B ^ 2 = root 3 times BC,
S △ ABC = 1 / 2acsinb, we can get that B ^ 2 = 6 times root 3, so B = 6 times root 3 under root

A question about the equal ratio sequence of high school mathematics
The sum of the first n terms of the sequence {the nth term of a} is SN. It is known that A1 = 1, and the (n + 1) term of a = Sn * (n + 2) / n
To prove: (1) the sequence {Sn / N} is an equal ratio sequence
(2) S (n + 1) = 4 * (the nth term of a)

Because a (n + 1) = (n + 2) / N * SN
So Sn = n * a (n + 1) / (n + 2)
S(n-1) = (n-1)*An / (n+1)
So an = Sn - S (n-1) = n / (n + 2) * a (n + 1) - (n-1) / (n + 1) * an
So 2n / (n + 1) * an = n / (n + 2) * a (n + 1)
That is, a (n + 1) / an = (2n + 4) / (n + 1)
So (Sn / N) / (s (n-1) / (n-1)) = (a (n + 1) / (n + 2)) / (an / (n + 1))
= A(n+1)/An * (n+1)/(n+2)
= (2n+4)/(n+1) * (n+1)/(n+2) = 2
So Sn / N is an equal ratio sequence with the common ratio of 2
(2)
Because Sn / N is an equal ratio sequence with 2 as the common ratio, the first term is S1 / 1 = S1 = A1 = 1
So the general formula of Sn / N is 2 ^ (n-1)
So Sn = n * 2 ^ (n-1)
S(n-1) = (n-1)*2^(n-2)
So an = Sn - S (n-1) = n * 2 ^ (n-1) - (n-1) * 2 ^ (n-2)
= n*2^(n-1) - n*2^(n-2) + 2^(n-2)
= n*2^(n-2) + 2^(n-2)
= (n+1) * 2^(n-2)
So the general formula is an = (n + 1) * 2 ^ (n-2)

Proof: 1 / (1 + 1)! + 2 / (2 + 1)! + +n/(n+1)!=1-1/(n+1)!

n/(n+1)!
=(n+1-1)/(n+1)!
=(n+1)/(n+1)!-1/(n+1)!
=
So the original formula = 1 / 1! - 1 / 2! + 1 / 2! - 1 / 3! + +1/n!-1/(n+1)!
=1-1/(n+1)!

Prove: (a ^ 1 / N + 1) / (n + 1) ^ 2 = 1)

According to Lagrange mean value theorem (a ^ 1 / n-a ^ 1 / (n + 1)) / (1 / N) - (1 / N + 1) = a ^ c * ina (C belongs to 1 / 1 + n to 1 / N)
）So (a ^ 1 / n-a ^ 1 / (n + 1)) / LNA = a ^ C / (n) * (n + 1) proves (a ^ 1 / N + 1) / (n + 1) ^ 2

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, S4 = 1, S8 = 17, and find the general term formula an

Let's set the public ratio of {an} {n} be q, which is known by S4 = 1, S8 = 17 to know Q ≠ 1, and get A1 (Q4 − 1) Q − 1 = 1, A1 (Q4 {4} 1 = 1) to get A1 (Q4 {1) Q {(Q44} 1 = 1) from S4 = 1, S8 = 1, S8 = 1, S8 = 1, S8 = 1, S8 = 1, and S8 = 1, and S8 = 1, and S8 = 1, and S8 = 1, and S8 − = (1) n × 2n {2n {n} 15, in this paper, we're talking about the fact that an = 2n {= 2n} 115 or an} 115 or an} 115 or an {115 or an {n}} 115 or an {n}

Let the sum of the first n terms of the sequence an be Sn, and Sn = n ^ 2-4n + 4, BN = an / 2 ^ n. can we use dislocation subtraction to find the sum of the first n terms of BN and TN?