1,3,6,10 . . n(n+1)/2 (通项） 这种数列的前19项和怎么求？
1,3,6,10 . . n(n+1)/2 (通项） 这种数列的前19项和怎么求？
1、数列求和的常用方法有：利用等差数列、等比数列的求和公式求和；分组求和；倒序相加；错位减法；裂项法。
2、本题中an=n(n+1)/2=(n²+n)/2，适宜用分组求和的方法求和。
3、解：因为1²+2²+3²+...+n²=n(n+1)(2n+1)/6，an=(n²+n)/2
所以a1+a2+...+a19=19*20*39/12+19(1+19)/4
=1330.
若数列an的通项为1/n(n+1)(n+2),求an的前n项和
an =1/[n(n+1)(n+2)]
=(1/2) {1/[n(n+1)] - 1/[(n+1)(n+2)] }
Sn = a1+a2+...+an
=(1/2) { 1/2 -1/[(n+1)(n+2)] }
数列AN通项公式AN=1/N(N+1),则数列 前10项和等于
解：
an=1/n(n+1)=1/n-1/(n+1)
S10=a1+a2+...a10
=1-1/2+1/2-1/3+....+1/10-1/11
=1-1/11
=10/11
a1=1,a n+1=(1+1/n)an+(n+1)/2^n. (1) 设bn=an/n,求数列（(bn)的通项公式 (2) 求数列（an)的前n项和Sn
1.a(n+1)=((n+1)/n)*an+(n+1)/2^n
a(n+1)/(n+1)=an/n+1/2^n
b(n+1)=bn+1/2^n
bn=b(n-1)+1/2^(n-1)
......
b2=b1+1/2^1
bn=b1+(1/2+1/2^2+...+1/2^(n-1))
bn=b1+1-1/2^(n-1)
bn=2-1/2^(n-1)
2.an=2n-n/2^(n-1)
sn=n(n+1)-(1+2/2+3/2^2+...+n/2^(n-1))
sn=n(n+1)-2*(1+2/2+3/2^2+...+n/2^(n-1)-1/2*(1+2/2+3/2^2+...+n/2^(n-1)))
sn=n(n+1)+2*(1+1/2+1/2^2+...+1/2^(n-1)-n/2^n)
sn=n(n+1)+2*(2-1/2^(n-1)-n/2^n)
sn=n(n+1)+4-(n+2)/2^(n-1)
数列 an 的前n项和sn，sn=n(n+1) 求an的通项
S(n-1)=(n-1)(n-1+1)=n(n-1)
an=Sn-S(n-1)=n(n+1)-n(n-1)=2n
已知数列{an}满足a1=2,a(n+1)=an-1/n(n+1）。（1）求数列的通项公式（2)设bn=nan*2^n,求数列的前n项和
(1)
a(n+1)=an -1/[n(n+1）]
a(n+1) - an = -1/[n(n+1)]
= 1/(n+1) - 1/n
an - a1 = 1/n -1
an = 1/n +1
(2)
bn = nan . 2^n
= n(1/n+1). 2^n
= (1+n) 2^n
= 2^n + n.2^n
summation bn
= summmation { (n+1)2^n}
consider
1+x+x^2+..+x^(n+1) = (x^(n+2)-1)/(x-1)
1+2x+..+(n+1) x^n
= [(x^(n+2)-1)/(x-1)]'
= [(n+1)x^(n+2) -(n+2)x^(n+1) +1]/(x-1)^2
2x+3x^2+..+(n+1)x^n
= [(n+1)x^(n+2) -(n+2)x^(n+1) +1]/(x-1)^2 -1
put x= 2
2.2^1+2.2^2+...(n+1).2^n
= (n+1)2^(n+2) - (n+2)2^(n+1) +1 -1
= n.2^(n+1)
summation bn
= n.2^(n+1)
数列{an}的前n项和Sn,2Sn=(n+1)an,1.且数列{an}的通项公式？2.求{1÷（an×an+1）}的前n项和Tn?
(1)2an=2Sn-2S(n-1)=(n+1)an-n(a(n-1))
所以an/a(n-1)=n/(n-1) (n≥2且n∈N*)
利用累乘法，得a1*a2/a1*a3/a2*……*an/a(n-1)=an=n (n∈N*)
所以{an}通项公式为an=n (n∈N*)
(2)新数列为1/n^2+1,即可求Tn
在数列{an}中，a1=1,an+1=(1+1/n)an+(n+1)/2^n(1)设bn=an/n求数列{bn}的通项公式（2）求数列{an}的前n项和
(1) ∵an+1=(1+1/n)an+(n+1)/2^n=(n+1)/n*an+(n+1)/2^n
∴an+1/(n+1)=an/n+1/2^n
∵bn=an/n，∴bn+1=bn+1/2^n
bn=bn-1+1/2^(n-1)
bn-1=bn-2+1/2^(n-2)
......
b2=b1+1/2^1
b1=a1/1=1
将上述n个式子加起来，得
bn=1+1/2+1/2^2+...+1/2^(n-1)
=1+1/2(1-(1/2)^(n-1))/(1-1/2)
=1+1-1/2^(n-1)
=2-1/2^(n-1)
(2) ∵bn=an/n，∴an=n*bn
∴Sn=1*b1+2*b2+...+n*bn
=1*(2-1)+2*(2-1/2)+3*(2-1/4)+...+n*(2-1/2^(n-1))
=2(1+2+3+...+n)-(1+2/2+3/4+...+n/2^(n-1))
=2Pn-Qn
设Pn=(1+2+3+...+n)=n(n+1)/2 Qn=(1+2/2+3/2^2+4/2^3+...+n/2^(n-1))
对Qn，有 Qn/2=(1/2+2/2^2+3/2^3+...+(n-1)/2^(n-1)+n/2^n)
Qn-Qn/2=Qn/2=1+[1/2+1/2^2+1/2^3+...+1/2^(n-1)]+n/2^n
=1+1/2(1-(1/2)^(n-1))/(1-1/2)+n/2^n
=1+1-(1/2)^(n-1)+n/2^n
=2-1/2^(n-1)+n/2^n
=2+(n+2)/2^n
∴Qn=4+2(n+2)/2^n
∴Sn=2Pn-Qn
=2*n(n+1)/2-4-2(n+2)/2^n
=n(n+1)-4-2(n+2)/2^n
设Sn为数列｛An｝的前n项和，a1＝3/2，a（n+1）＝2Sn+4n。（1）求数列｛An｝的通项公式
a(n+1)-an=2Sn+4n-2S(n-1)-4(n-1)=2an+4
a(n+1)=3an+4
a(n+1)+2=3an+6=3(an+2)
所以数列{an+2}是等比数列，公比q=3，首项a1+2=3/2+2=7/2
所以an+2=7/2*3^(n-1)
an=7/2*3^(n-1)-2
