已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn？
已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn？
an = (3n-2).3^(n+1)
= 9(n.3^n) -2.3^(n+1)
Sn = an+a2+...+an
=9[∑(i:1->n)i.3^i ] - 9(3^n-1)
let
S = 1.3 + 2.3^2 ...+n.3^n (1)
3S = 1.3^2+2.3^3+...+n.3^(n+1) (2)
(2)-(1)
2S =n.3^(n+1) -( 3+3^2+...+3^n)
=n.3^(n+1) - (3/2)(3^n-1)
S = (1/2)[n.3^(n+1) - (3/2)(3^n-1)]
Sn = an+a2+...+an
=9[∑(i:1->n)i.3^i ] - 9(3^n-1)
=9S -9(3^n-1)
= (9/2)[n.3^(n+1) - (3/2)(3^n-1)] -9(3^n-1)
= (9/2)n.3^(n+1) - (7/4) .3^(n+2) + 9
已知数列{an}中,sn=2的n+1次方-2
sn=2（2^n-1），首项a1=2,公比2的等比数列
an=2^n
已知数列an中,an=2n-3/3^n+1,求sn
an=(2n-3)/3^(n+1)=2n/3^(n+1) -1/3ⁿ
Sn=a1+a2+...+an
=2[1/3²+2/3³+...+n/3^(n+1)] -(1/3+1/3²+...+1/3ⁿ)
令Cn=1/3²+2/3³+3/3⁴+...+n/3^(n+1)
则Cn /3=1/3³+2/3⁴+...+(n-1)/3^(n+1)+n/3^(n+2)
Cn-Cn/3=(2/3)Cn=1/3²+1/3³+...+1/3^(n+1) -n/3^(n+2)
=(1/9)×(1-1/3ⁿ)/(1-1/3) -n/3^(n+2)
=(1/6)×(1-1/3ⁿ) -n/3^(n+2)
Cn=(1/4)(1- 1/3ⁿ) -n/[2×3^(n+1)]
Sn=2Cn -(1/3+1/3²+...+1/3ⁿ)
=-n/ 3^(n+1)
已知数列an=（3n-1）*（2/3）的n-1次方，求Sn
Sn=2+5*(2/3)^1+8*(2/3)^2+……+(3n-1)*(2/3)^(n-1) (1)(2/3)Sn=2*(2/3)^1+5*(2/3)^2+8*(2/3)^3+……+(3n-1)*(2/3)^n (2)(2)-(1)得-(1/3)Sn=-2-3*(2/3)^1-3*(2/3)^2-3*(2/3)^3-……-3*(2/3)^(n-1)+(3n-1)*(2/3)^n =-2-[3*(2/3)^1+3*(2/3)^2+3*(2/3)^3+……+3*(2/3)^(n-1)]+(3n-1)*(2/3)^n =-2-3[1-(2/3)^（n-1）)/[1-(2/3)]+(3n-1)*(2/3)^n sn=15-(9n+33/2)*(2/3)^n
已知数列an中,a1=3,a（n+1）=3an+2,求Sn
a（n+1）+1=3(an+1)
设bn=an+1,则bn是以b1=4,q=3的等比数列。Tn=b1(1-qn)/(1-q)=2x3^n-2
又Tn=Sn+n,故，Sn=Tn-n=2x3^n-n-2
已知数列{an}通项an=(2n-1)*3^n+1,求Sn
解答：错位相减
Sn =1*3^2+3*3^3+5*3^4+........+(2n-3)*3^n+(2n-1)*3^(n+1) (1)
同乘以3
3Sn = 1*3^3+3*3^4+..............................+(2n-3)*3^(n+1)+(2n-1)*3^(n+2) (2)
(1)-(2)
-2Sn =3^2+2[3^3+3^4+.......................................+3^(n+1)] -(2n-1)*3^(n+2)
-2Sn=9+2*[27-3^(n+2)]/(1-3)-(2n-1)*3^(n+2)
-2Sn=9+3^(n+2)-27-(2n-1)*3^(n+2)
Sn=9+(n-1)*3^(n+2)
已知数列{an}满足a1=1，an+a（n+1）=（1/3）的n次方（n∈N），Sn=a1+3a2+3²a3……+3n-1次方an
由Sn=a1+3a2+3²a3……+3n-1次方an 得
3Sn= 3a1+3²a2……+3n-1次方an-1+3n次方×an
相加4Sn=a1+3(a1+a2)+……3n-1次方*(an-1+an)+3n次方an=a1+(n-1)+3n次方×an
故：4Sn-3的n次方×an=a1+n-1=n
已知数列{an}满足:a1=2,an+1=3an+3的n+1次方-2的n次方(n∈N+)
a(n+1)=3an+3^(n+1)-2^n
a(n+1)/3^(n+1) - an/3^n = -(1/3)(2/3)^n
an/3^n - a(n-1)/3^(n-1) = -(1/3)(2/3)^(n-1)
an/3^n - a1/3 = -[ 1-(2/3)^n]
an/3^n = -1/3+ (2/3)^n
an = 2^n - 3^(n-1)
= an + 1/an
let
f(x) = 2^x -3^(x-1) + 1/[2^x -3^(x-1)]
f'(x)=(ln2).2^x - (ln3). 3^(x-1) - [ (ln2).2^x - (ln3). 3^(x-1) ]/1/[2^x -3^(x-1)]^2
f'(x) <=0 ( for x>=1 )
f(1) = 2
f(2) =2
f(3) = -2
{} 是递减数列
ie
k =1 or 2
<=ck
已知数列an=2×3的n-1次方+(-1)的n次方×(ln2-ln3)+(-1)的n次方×n×ln3 求Sn
a（n+1）=2×3的n次方+(-1)的n+1次方×(ln2-ln3)+(-1)的n+1次方×(n+1)×ln3
这个等式加去原来等式：当n为奇数：a（n+1）+an=6^n+6^(n-1)+ln3于是6a（n+1）-6^(n+1)+6an-6^n=6ln3,a(1)=ln2
换元t（n+1）+t（n）=6ln3，t（1）=6ln2-6
t（2）=6ln3-6ln2+6，t（3）=6ln2-6这个数列回圈，所以t（n）=6ln2-6：n为奇数
t（n）=6ln3-6ln2+6：n为偶
得6an-6^n=6ln2-6：n为奇数
做到这步，求大神来帮下
已知数列an,an=3n-4+2的n次方分之1，则Sn=
Sn=3*1-4+1/2^1+3*2-4+1/2^2+3*3-4+1/2^3+..........+3*n-4+1/2^n
=(3*1-4+3*2-4+3*3-4+........+3*n-4)+(1/2^1+1/2^2+1/2^3+........+1/2^n)
=(-1+3*n-4)n/2+1/2*[1-(1/2)^n]/(1-1/2)
=n(3n-5)/2+1-(1/2)^n
