已知等差数列{an}的前n项和是Sn,且a1=2,S5=30
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已知等差数列{an}的前n项和是Sn,且a1=2,S5=30
已知等差数列{an}的前n项和是Sn,且a1=2,S5=30(1)求数列{an}的通项公式 (2)若数列{bn}满足bn=2的an-1次方,求数列{bn}的前n项和是Tn (3)若数列{Cn}满足Cn=1/Sn,求数列{Cn}的前n项和是Rn
已知等差数列{an}的前n项和是Sn,且a1=2,S5=30(1)求数列{an}的通项公式 (2)若数列{bn}满足bn=2的an-1次方,求数列{bn}的前n项和是Tn (3)若数列{Cn}满足Cn=1/Sn,求数列{Cn}的前n项和是Rn
![已知等差数列{an}的前n项和是Sn,且a1=2,S5=30](/uploads/image/z/15928037-53-7.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E6%98%AFSn%2C%E4%B8%94a1%3D2%2CS5%3D30)
(1)
an=a1+(n-1)d= 2+(n-1)d
S5 = (2+2d)5 = 30
d=2
an = 2+(n-1)2 = 2n
(2)
bn=2^(an-1)
= 2^(2n-1)
Tn =b1+b2+...+bn
= 2(2^(2n) -1) / (4-1)
= (2/3)(2^(2n) -1)
(3)
cn =1/Sn
= 1/ [n(n+1)]
= 1/n -1/(n+1)
Rn = c1+c2+...+cn
= 1-1/(n+1)
= n/(n+1)
an=a1+(n-1)d= 2+(n-1)d
S5 = (2+2d)5 = 30
d=2
an = 2+(n-1)2 = 2n
(2)
bn=2^(an-1)
= 2^(2n-1)
Tn =b1+b2+...+bn
= 2(2^(2n) -1) / (4-1)
= (2/3)(2^(2n) -1)
(3)
cn =1/Sn
= 1/ [n(n+1)]
= 1/n -1/(n+1)
Rn = c1+c2+...+cn
= 1-1/(n+1)
= n/(n+1)
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