等差数列{Sn}中,a1=1,前n项和Sn满足条件 S2n/Sn=4,n=1,2,.
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等差数列{Sn}中,a1=1,前n项和Sn满足条件 S2n/Sn=4,n=1,2,.
(1)求数列{an}的通项公式an和前n项和Sn:
(2)记bn=an*2^(n-1),求数列{bn}前n项和Tn:
(1)求数列{an}的通项公式an和前n项和Sn:
(2)记bn=an*2^(n-1),求数列{bn}前n项和Tn:
(1)S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)
Sn = n(a1+an)/2
S(2n)/Sn = 2(a1+a2n)/(a1+an) = 4
a1+a2n =2a1+2an
a2n = an + nd,其中d为等差公比;
∴an +nd = a1 + 2an
an = nd - a1
而:an = a1 + (n-1)d
对比:d = 2
所以:an = 2n-1
Sn = n^2
(2)bn = an*2^(n-1) = (2n-1) *2^(n-1)
Tn = 2^0 + 3*2^1 + 5 * 2^2 + 7*2^3+.+(2n-1) *2^(n-1)
2Tn =2 + 3**2^2+5 * 2^3+.+(2n-3)*2^(n-1)+(2n-1) *2^n
两式相减:
-Tn = 2^0 + 2*2 + 2*2^2 +.+2^2^(n-1)-(2n-1) *2^n
Tn = -1+(2n-1) *2^n - 4(2^n -1)
再问: 为什么S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)?
Sn = n(a1+an)/2
S(2n)/Sn = 2(a1+a2n)/(a1+an) = 4
a1+a2n =2a1+2an
a2n = an + nd,其中d为等差公比;
∴an +nd = a1 + 2an
an = nd - a1
而:an = a1 + (n-1)d
对比:d = 2
所以:an = 2n-1
Sn = n^2
(2)bn = an*2^(n-1) = (2n-1) *2^(n-1)
Tn = 2^0 + 3*2^1 + 5 * 2^2 + 7*2^3+.+(2n-1) *2^(n-1)
2Tn =2 + 3**2^2+5 * 2^3+.+(2n-3)*2^(n-1)+(2n-1) *2^n
两式相减:
-Tn = 2^0 + 2*2 + 2*2^2 +.+2^2^(n-1)-(2n-1) *2^n
Tn = -1+(2n-1) *2^n - 4(2^n -1)
再问: 为什么S(2n) = 2n(a1+a2n)/2 = n(a1+a2n)?
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