If we know the leading formula of sequence an an = n Λ 2cosn π and Sn is the sum of its first n terms, then s2010 / 2010=

If we know the leading formula of sequence an an = n Λ 2cosn π and Sn is the sum of its first n terms, then s2010 / 2010=

S2010/2010=(a1+a2+a3+a4+… +A2010) / 2010 = (- 1 + 2 square - 3 square + 4 square +...) +2010 Square) / 2010 = (1 + 2 + 3 + 4 + +2010)/2010=(2011*2010/2)/2010=2011/2

The general term formula of sequence {an} is an = ncosn π 2 + 1, the sum of the first n terms is Sn, then s2012=______ ．

Because Cosn π 2 = 0, - 1, 0, 1, 0, - 1, 0, 1 ；∴ncosnπ2=0，-2，0，4，0，-6，0，8… The sum of every four terms of the sequence {an} is 2, while the sum of every four terms of the sequence {an} is 2 + 4 = 6, while the sum of every four terms of the sequence {an} is 2 + 4 = 6, and the sum of every four terms of the sequence {an} is 2 + 4 = 6

If the first n terms of sequence {an} and Sn = N2 + 2n + 5, then A6 + A7 + A8=

a6+a7+a8
=S8-S5
=8²+2×8+5-(5²+2×5+5)
=45

Let the first n terms of sequence {an} and Sn = N2, then A8=______ ．

∵ an = sn-sn-1 (n ≥ 2), Sn = N2 ∵ A8 = s8-s7 = 64-49 = 15, so the answer is 15

If the general term of the sequence {an} is an = N2 (cos2n π 3-sin2n π 3), and the sum of the first n terms is Sn, then S30 is______ ．

∵an=n2（cos2nπ3-sin2nπ3）=n2cos2nπ3∴S30＝12•cos2π3+22cos4π3+32cos2π+… +302cos20π=−12×1−12×22+32−12×42−12×52+62+… −12×282−12×292+302=−12[1+22-2×32）+（42+52-62×2）+… +（282+292...

If the general term of the sequence {an} is an = (- 1) ^ n * n * sin (n π / 2) + 1, and the sum of the first n terms is Sn, then S100 =? Urgent!

an = 1+ nsin(nπ/2).(-1)^n
a1= 1-1=0
a2= 0
a3= 1+3=4
a4= 0
ie
an= 1- n ; n=4m-3
= 1+n ; n=4m-1
= 0 ; n=4m-2 or 4m
S100=a1+a2+...+a100
= a1+a3+.+a99
=(1-1) +(1+3) +(1-5)-(1+7) +.+(1+99)
= 0 +4 +(-4)+.+(-96)+100
= 100

It is known that each item of the sequence {an} is a positive number, the sum of the first n items is Sn, and Sn = an (an + 1) 2, n ∈ n +

It is proved that: SN = an (an + 1) 2 ∫ S1 = A1 (1 + A1) 2 ∫ A1 = 1 (1 point) from 2Sn = A2N + an2sn-1 = a2n-1 + an-1 {2An = 2 (sn-sn-1) = a2n-a2n-1 + an-an-1 (3 points) so (an + an-1) (an-an-1-1) = 0 ∵ an + an-1 ﹥ an-an-1 = 1, so the sequence {an} is an arithmetic sequence (6 points)

Let the first n terms and Sn of the equal ratio sequence {an}, if Sn = 3 ^ Na + B, and a ≠ 0, a and B are constants, then a + B=
In the sequence {an}, if A1 = 2, Anan + 1 + an + 1 + 1 = 0, then s2010=

a1=S1=a+b
n> 1, an = SN-S (n-1) = a * 3 ^ n + B - [a * 3 ^ (n-1) + b] = 2A * 3 ^ (n-1)
A2 = 6a, equal ratio q = 3
So 3A1 = A23 (a + b) = 6a, a = B
The sum of the first n terms is A1 (3 ^ n-1) / 2 = a * 3 ^ n-a
We get a = b = 0
So a + B = 0

The arithmetic sequence {an}, A5 = 4, a7 = 6, an =? Sn?

A7 / A5 = q ^ 2, so q = √ 6 / 2, A1 = 16 / 9, an = A1 * q ^ (n-1) = 16 / 9 * (√ 6 / 2) ^ (n-1)
sn=a1(1-q^n)/1-q=-32(2+√6)[1-(√6/2)^n]/9

Given an = (2n + 1) * 3 ^ n, find SN

an=(2n+1)*3^n
a1=3*3^1
a2=5*3^2
a3=7*3^3
.
an=(2n+1)*3^n
Sn=3*3^1+5*3^2+7*3^3+.(2n+1)*3^n
3Sn= 3*3^2+5*3^3+7*3^4+.+(2n-1)*3^n+(2n+1)*3^(n+1)
3Sn-Sn=2Sn=-3*3-2*3^2-2*3^3-2*3^4-.-2*3^n+(2n+1)*3^(n+1)
2Sn=-9-2(3^2+3^3+3^4+.+3^n)+(2n+1)*3^(n+1)
2Sn=-9-2*3^2(1-3^(n-1))/(1-3)+(2n+1)*3^(n+1)
2Sn=-9+3^2(1-3^(n-1))+(2n+1)*3^(n+1)
2Sn=-3^(n+1)+(2n+1)*3^(n+1)
2Sn=2n3^(n+1)
Sn=n3^(n+1)
If you agree with my answer, please click "adopt as a satisfactory answer", thank you!