Question

已知等差数列｛an｝满足：a3=7，a5+a7=22.

已知等差数列｛an｝满足：a3=7，a5+a7=22.｛an｝的前n项和为Sn。
1.求an及sn
2. 求令bn=1/(anan+1)，求｛bn｝及前n项和Tn

Answer 1

1、∵数列an是等差数列
∴a5+a7=a3+2d+a3+4d
=2a3+6d
=14+6d
=22
即d=4/3，a1=a3-2d=13/3
an=a1+(n-1)d=13/3+(n-1)*(4/3)=4n/3+3
Sn=(a1+an)n/2=(4/3+4n/3+3)n/2=2n²/3+13n/6

2、bn=1/(anan+1)
=1/[(4n/3+3)(4n/3+13/3)]
=1/{[(4n+9)/3]*[(4n+13)/3]}
=9/[(4n+9)(4n+13)]
=9/4*[1/(4n+9)-1/(4n+13)]
Tn=b1+b2+……+bn
=9/4*(1/13-1/17)+3/4*(1/17-1/21)+……+3/4*[1/(4n+9)-1/(4n+13)]
=9/4*[1/13-1/17+1/17-1/21+……+1/(4n+9)-1/(4n+13)]
=9/4*[1/13-1/(4n+13)]
=9/4*{(4n+13-13)/[13(4n+13)]}
=9/4*{4n/[13(4n+13)]}
=9n/(52n+169)

Answer 2

解：∵{an}是等差数列，
∴a3=a1+2d=7 ①
a5=a1+4d
a7=a1+6d
又a5+a7=22
∴2a1+10d=22 ②
联立①②形成方程组并解知得：
a1=13/3，d=4/3
∴an=a1+(n-1)d
=13/3+4n/3-4/3
=3+4n/3
Sn=(na1)+[n(n-1）]d/2
=(13n/3)+[4n(n-1)/6]
又bn=1/(anan+1)
∴b1=1/[(13/3)^2+1)
=9/[(13)^2+3^2]
b2=1/{[(13/3)+(4/3）]^2+1}
=9/[(17)^2+3^2]
b3=1/{[(13/3)+2*4/3]^2+1}
=9/[(21)+3^2]
……
bn=9/{[(9+4n)^2+3^2]

Answer 3

a(n)=a+(n-1)d,
7=a(3)=a+2d,
22=a(5)+a(7)=2a+10d, 11=a+5d=a+2d+3d=7+3d, d=4/3, a=7-2d=13/3
a(n)=13/3 +4(n-1)/3,
s(n)=13n/3 + 2n(n-1)/3,

a(n+1)=13/3 + 4n/3,
b(n)=1/[a(n)a(n+1)] = 1/[(13/3+4(n-1)/3)(13/3+4n/3)]=9/[(9+4n)(13+4n)]
=(9/4)[1/(9+4n) - 1/(13+4n)]
=(9/4)[1/(9+4n) - 1/[9+4(n+1)] ]
t(n)=b(1)+b(2)+...+b(n-1)+b(n)
=(9/4)[1/13-1/17 + 1/17-1/21 + ... + 1/(9+4n-4) - 1/(9+4n) + 1/(9+4n)-1/(9+4n+4)]
=(9/4)[1/13 - 1/(9+4n+4)]
=(9/4)[1/13 - 1/(13+4n)]
=9n/[13(13+4n)]

Answer 4

1.求an及sn

a3=7，a5+a7=22

a3+a5+a7=3a5=22+7=29

a5=29/3

公差B=(a5-a3)/2=(29/3-7)/2=4/3
a0=7-12/3=3

an=3+4/3n