已知等差数列(an)满足:a3=7,a5+a7=26,(an)的前n项和为Sn
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/08/17 15:32:13
已知等差数列(an)满足:a3=7,a5+a7=26,(an)的前n项和为Sn
(1)求an及sn
(2)令bn=an平方-1分之1(n属于n)求数列(bn)的前n项和Tn
(1)求an及sn
(2)令bn=an平方-1分之1(n属于n)求数列(bn)的前n项和Tn
![已知等差数列(an)满足:a3=7,a5+a7=26,(an)的前n项和为Sn](/uploads/image/z/5084205-69-5.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%88an%EF%BC%89%E6%BB%A1%E8%B6%B3%EF%BC%9Aa3%3D7%2Ca5%2Ba7%3D26%2C%EF%BC%88an%EF%BC%89%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn)
/>(1)设等差数列的首项是a1,公差为d
则 a1+2d=7 ①
2a1+10d=26 ②
①*5-②
解得 a1=3,∴ d=2
∴ an=a1+(n-1)d=2n+1
∴ Sn=(a1+an)*n/2=(2n+4)*n/2=n(n+2)
(2)bn=an平方-1分之1
=1/[(2n+1)²-1]
=1/[2n(2n+2)]
=(1/4)*1/[n(n+1)]
=(1/4)*[1/n-1/(n+1)]
∴ 数列(bn)的前n项和Tn为:
Tn=(1/4)*[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=(1/4)*[1-1/(n+1)]
=(1/4)*n/(n+1)
则 a1+2d=7 ①
2a1+10d=26 ②
①*5-②
解得 a1=3,∴ d=2
∴ an=a1+(n-1)d=2n+1
∴ Sn=(a1+an)*n/2=(2n+4)*n/2=n(n+2)
(2)bn=an平方-1分之1
=1/[(2n+1)²-1]
=1/[2n(2n+2)]
=(1/4)*1/[n(n+1)]
=(1/4)*[1/n-1/(n+1)]
∴ 数列(bn)的前n项和Tn为:
Tn=(1/4)*[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=(1/4)*[1-1/(n+1)]
=(1/4)*n/(n+1)
已知等差数列{An}满足:a3=7 ,a5+a7=26 ,{An}的前n项和为Sn
已知等差数列{an}满足:a3=7,a5+a7 =26,an的前n项和为sn.
已知等差数列(an)满足:a3=7,a5+a7=26,(an)的前n项和为Sn
已知等差数列{an}满足:a3=7,a5+a7=26.{an}的前n项和为Sn.
已知等差数列an满足a3=7,a5+a7=26,an的前n项和为sn.1.求an和sn
已知等差数列{an}满足:a3=7,a5+a7=26,{an}的前n项和为Sn.求an及Sn.
已知等差数列{an}满足a3=7,a5+a7=26{an}的前n项和为Sn求a4及Sn
1.已知等差数列{An}满足:A3=7,A5+A7=26,{An}的前n项和为Tn
已知等差数列{an}满足:a3=7,a5+a7=26,{an}的前n项和为Sn.令bn=1/(an)^2-1(n∈N*)
已知等差数列【an】满足:a3=7,a5+a7=26.【an】的前n项和为Sn.(1)求a4及Sn;
1.已知等差数列{an}满足a3=7,a5+a7=26,数列的前n项和为Sn令bn=1/(an²-1) &nb
已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和T