Let an = (3n-2) multiply by 3 to the nth power, and find Sn (dislocation subtraction) in detail. Please write down the step that 3 times Sn minus Sn =? Is equal to finally

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An = (3n-5) * 2 ^ n, find the sum of the first n terms, Sn, and subtract by dislocation

Sn=-2*2^1+1*2^2+4*2^3+7*2^4+...+(3n-5)*2^n,
Sn/2=-2+1*2^1+4*2^2+7*2^3+...+(3n-5)*2^(n-1),
The result of subtracting the two formulas is as follows:
Sn/2=2-3[2+2^2+2^3+...+2^(n-1)]+(3n-5)*2^n,
=2-3*2*[1-2^(n-1)]/(1-2)+(3n-5)*2^n
=8-(3n-8)*2^n,
——》Sn=16-(3n-8)*2^(n+1).

What is 1-3x = 1.5x-0.75x

7/18

Given the function f (x) = 3x / x + 3 (x is not equal to negative 3, X belongs to R), the sequence {a small n} satisfies a small n = f (a small n minus 1) (n is greater than or equal to 2, n belongs to n) and A1 is unequal
Given the function f (x) = 3x / x + 3 (x is not equal to negative 3, X belongs to R), the sequence {a small n} satisfies a small n = f (a small n minus 1) (n is greater than or equal to 2, n belongs to n) and A1 is not equal to 0, (1) prove that the sequence {1 / a small n} is the arithmetic sequence (2). If A1 = 1 / 4, find the value of A5

(1) Let BN = 1 / an
Bn - B(n-1)=1/f(A(n-1))-1/f(A(n-1))
=(f(A(n-1)) +3)/(3f(A(n-1)) ) - 1/f(A(n-1))
=1/3+1/f(A(n-1)) -1/f(A(n-1))
=1/3
(2)A1=0.25 B1=4
BN is an arithmetic sequence, BN = n / 3 + 11 / 3
So an = 3 / (n + 11)

Given that the sum of the first n terms of a (n) is s (n), a (n) is equal to 1 / 2 of (2n-1) (4N + 2), find s (n)?

an=1/(2n-1)(4n+2)=1/2*1/(2n-1)(2n+1)=1/2*1/2*[1/(2n-1)-1/(2n+1)]
Sn=1/4(1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1))
=1/4(1-1/(2n+1))
=1/4*2n/(2n+1)
=n/[2(2n+1)]

Sequence 3, 4, 5 The sum of the first n terms of (n + 1) is [3 + (n + 1)] (n-1) / 2, right

Sequence 3, 4, 5 (n+2)
An = n + 2, not equal to N + 1
When tolerance d = 1: SN = (a1 + an) n / 2 = [3 + (n + 2)] / 2 = 3

If the sum of the first n terms of an is known, Sn = 2 ^ n-a, then A1 ^ 2 + A2 ^ 2 + a3 ^ 2 +. + an ^ 2 is equal to

If Sn = 2 ^ n - A is known, then:
S(n-1)=2^(n-1) - a
By subtracting the two formulas, we get the following result:
an=2^n - 2^(n-1) =2^(n-1)
a1=2^(1-1) =1
a2=2^(2-1)=2
a3=2^(3-1)=4
a4=2^(4-1)=8.
Therefore, the sequence {an} is an equal ratio sequence with the first term of 1 and the common ratio of 2
Then:
a1^2=1
a2^2=4
a3^2=16
a4^2=64.
Therefore, A1 ^ 2 + A2 ^ 2 +... + an ^ 2 can be regarded as the sum of the equal ratio sequence with the first term of 1 and the common ratio of 4, then: A1 ^ 2 + A2 ^ 2 +. + an ^ 2 = (1-4 & # 8319;) / (1-4) = (4 & # 8319; - 1) / 3

It is known that a1 + A2 + a3 = 8, a1 + A2 + +a6=7,Sn=a1+a2+a3+… +An, then SN is equal to?
To specific problem-solving process, to speed up oh

Let the common ratio of a sequence be Q
There are
a1+a2+a3=a1(1+q+q^2)=8
a1+a2+… +a6=a1(1+q+q^2+q^3+q^4+q^5)=7
Divide the second by the first, and you get
1 + Q ^ 3 = 7 / 8, q = 1 / 2
Substituting into the first formula, the solution is A1 = 32 / 7
therefore
Sn=a1(1-q^n)/1-q
=(64/7)*(1-(1/2)^n)

In {an}, Sn = k - (1 / 2) ^ 2, then the value of real number k is

Change the title to
In {an}, Sn = k - (1 / 2) ^ n, then the value of real number k is
a1=s1=k-1/2,
a1+a2=s2=k-1/4,
∴a2=1/4,
a1+a2+a3=s3=k-1/8,
∴a3=1/8,
From A1 * A3 = A2 ^ 2, (k-1 / 2) / 8 = 1 / 16,
∴k=1.

Given the first n terms and Sn = 2an-2n of sequence an, prove that sequence (an + 1-2an) is arithmetic sequence 2. Prove that (an + 2) is proportional sequence 3. Find the general formula of an?

It is proved that: 1. S (n + 1) = 2A (n + 1) - 2 (n + 1) a (n + 1) = s (n + 1) - Sn a (n + 1) - 2An = 2 is equal difference sequence 2. From above, a (n + 1) + 2 = 2 (an + 2) is obtained. Therefore: {an + 2} common ratio is 2, equal ratio sequence A1 = 2
3. An + 2 = (a1 + 2) 2 ^ (n-1) so: an = 2 ^ (n + 1) - 2

Let an = (3n-2) multiply by 3 to the nth power, and find Sn (dislocation subtraction) in detail. Please write down the step that 3 times Sn minus Sn =? Is equal to finally