In the sequence {an}, if A1 = 2 / 3, then a (n + 1) = 1 / (n + 1) (n + 2) is equal to a(n+1)=1/(n+1)(n+2)+an

In the sequence {an}, if A1 = 2 / 3, then a (n + 1) = 1 / (n + 1) (n + 2) is equal to a(n+1)=1/(n+1)(n+2)+an

Because a (n + 1) = 1 / (n + 1) (n + 2) + an
So a (n + 1) - an = 1 / (n + 1) (n + 2) = 1 / (n + 1) - 1 / (n + 2) (split term)
So a2-a1 = 1 / 2-1 / 3
a3-a2=1/3-1/4
a4-a3=1/4-1/5
.
an-a(n-1)=1/n-1/(n+1)
Superimposed
an-a1=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...+(1/n-1/(n+1))=1/2-1/(n+1)
Because A1 = 2 / 3
So an = a1 + 1 / 2-1 / (n + 1) = 2 / 3 + 1 / 2-1 / (n + 1) = 7 / 6-1 / (n + 1)

In the sequence {an}, if A1 = 2, Nan + 1 = (n + 1) an + 2 (n ∈ n *), then A10 is ()
A. 34B. 36C. 38D. 40

∵nan+1=（n+1）an+2∴an+1n+1−ann＝2n(n+1)＝2(1n −1n+1)∴a1010＝a1010−a99+a99−a88+… +a22−a11+a1=2[（19−110）+（18−19）+… +(1-12)] + 2 = 3810a10 = 38, so C

It is known that in the sequence {an}, A1 = 2, Na [n + 1] = (n + 1) an + 2 (n ∈ n *), A10 = () Note: [] is a subscript

Divide the two sides by n (n + 1) a (n + 1) / (n + 1) = an / N + 2 / N (n + 1) a (n + 1) / (n + 1) - an / N = 2 / N (n + 1) = 2 [1 / n-1 / (n + 1)] A10 / 10-a9 / 9 = 2 (1 / 10-1 / 11) A9 / 9-a8 / 8 = 2 (1 / 9-1 / 10) A2 / 2-a1 / 1 = 2 (1 / 2-1 / 3) add A10 / 10-a1 / 1 = 2 * (1 / 2-1 / 11) = 9 / 11a10 = 310 / 11

If the sum of the first n terms of the sequence {an} is Sn, and an = Sn × s (n-1) (n ≥ 2), A1 = 2 / 9, then what is A10 equal to

An = Sn × s (n-1) (n ≥ 2) A1 = 2 / 9, S1 = 2 / 9A2 = S2 * S1 = (2 / 9 + A2 *) 2 / 9 get A2 = 4 / 63, S2 = a1 + A2 = 2 / 9 + 4 / 63 = 18 / 63 = 2 / 7An = Sn × s (n-1) SN-S (n-1) = Sn × s (n-1) get 1 / S (n-1) - 1 / Sn = 1

Given a (n + 1) = Nan + n-1, A1 = 1, find the general term formula of sequence {an}

a(n+1)=nan+n-1,
a(n+1)+1=nan+n,
(a(n+1)+1)/(an+1)=n,
So an + 1 = (an + 1) / (a (n-1) + 1) & _; (a (n-1) + 1) / (a (n-2) + 1)
•…… •(a3+1)/(a2+1)•(a2+1)/(a1+1)•(a1+1)
=(n-1) •(n-2) •…… •2•1•2=2•(n-1)!
∴an=2•(n-1)!-1.

It is known that the sum of the first n terms of the equal ratio sequence {an} is 3 ^ (- n) - C. the prime of the positive term sequence {BN} is C, and the first n terms of the sequence and Sn satisfy the following conditions
Sn - √ Sn = s (n-1) + √ s (n-1), (n ≥ 2)
(1) Find C, and find the general formula of sequence {an} and {BN};
(2) Find the sum of the first n terms of the sequence {BN times (1-1 / 2 times an} as TN;

(1) Sn - √ Sn = s (n-1) + √ s (n-1), SN-S (n-1) = √ Sn + √ s (n-1), Sn is not equal to s (n-1),
Divide both sides by √ Sn + √ s (n-1) to get √ Sn - √ s (n-1) = 1, so {Sn} is an arithmetic sequence with the first term of C and the tolerance of 1
√Sn=c+n-1,bn=Sn-S(n-1)=2c+2n-3
And B1 = C = 2c-1, so C = 1, BN = 2N-1
Bn=3^(-n) -1,an=Tn-T(n-1)=(-2)*3^(-n)
(2)Tn=(2n-1)*[1+3^(-n)]

Given the sum of the first n terms of the sequence {an} and Sn = 12n-n ^ 2, what is the sum of the first n terms of the sequence with an absolute value

Sn=12-n²
an＝Sn-S(n-1)＝13-2n
It's a decreasing sequence
Let an6.5, that is, the first six items are positive, and then negative!
Therefore, the sum of the first n terms is as follows:
(1) When n ≤ 6
Sn=12n-n²
(2) When n ≥ 7
|a1+|a2|+|a3|+…… +||an|
＝S6-(a7+a8+…… +an)
＝2S6-(a1+a2+…… +a6+a7+…… +an)
＝2S6-Sn＝2*(12*6-6²)+n²-12n
＝n²-12n+72

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, A2 equals 3, S6 equals 36
It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, A2 is equal to 3, S6 is equal to 36

S6=3(a1+a6)=3(a2+a5)
36=3(3+a5)
a5=9
3d=a5-a2=9-3=6
d=2
a1=a2-d=1
an=a1+(n-1)d=1+2(n-1)=2n-1
an/2^n=(2n-1)/2^n
Tn=a1/2+a2/2^2+… +an/2^n
1/2Tn= a1/2^2+… +a(n-1)/2^n+an/2^(n+1)
T n = 3 - (2n + 3) / 2 ^ n obtained by dislocation subtraction method

The sum of the first n terms of the sequence an is Sn, and satisfies an + 2Sn * s (n-1) = 0, (n is greater than or equal to 2), A1 = 1 / 2. Prove the 1 / Sn equal difference, find the expression of an
The sum of the first n terms of the sequence an is Sn, and satisfies an + 2Sn * s (n-1) = 0, (n is greater than or equal to 2), A1 = 1 / 2. Prove 1 / Sn equal difference, find an expression!

an+2Sn*S(n-1)=0
And an = SN-S (n-1)
∴Sn-S(n-1)+2Sn*S(n-1)＝0
Same as Sn * s (n-1)
1/Sn -1/S(n-1)＝2
{1 / Sn} is an arithmetic sequence, tolerance 2, first term = 1 / A1 = 2
1/Sn=2+2(n-1)=2n
Sn=1/(2n)
an=Sn-S(n-1)=1/(2n)-1/(2n-2)

In known sequence {an}, A1 = 3, an + 1 = 2 (a1 + A2 + a3. + an), then the general term formula of sequence is given

n> N = 2: an + 1 = 2 (a1 + A2 + a3. + an) = 2Sn, so Sn = 1 / 2An + 1
an=...=2Sn-1.Sn-1=1/2an.
Sn-sn-1 = an = 1 / 2An + 1-1 / 2An, so an + 1 = 3an
The common ratio of an ^ = 2.3 (n = 2.3)
n=1,a1=3