已知{an}是首项为19，公差为-2的等差数列，Sn为{An}的前n项和。(1)求通项an及前n项和Sn(2)设｛bn-an｝
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(1)
a1=19, d=-2
an = 19-2(n-1) = 21-2n
Sn = (21-2n+19)n/2 = (20-n)n
(2)
(bn-an) = (b1-a1).q^(n-1)
= 3^(n-1)
bn = an + 3^(n-1)
=21-2n + 3^(n-1)
Tn = b1+b2+...+bn
= 21n(n+1)/2 - (1/3)n(n+1)(2n+1) + (1/2)(3^n-1)
= (1/6)n(n+1)(61-4n) +(1/2)(3^n-1)
a1=19, d=-2
an = 19-2(n-1) = 21-2n
Sn = (21-2n+19)n/2 = (20-n)n
(2)
(bn-an) = (b1-a1).q^(n-1)
= 3^(n-1)
bn = an + 3^(n-1)
=21-2n + 3^(n-1)
Tn = b1+b2+...+bn
= 21n(n+1)/2 - (1/3)n(n+1)(2n+1) + (1/2)(3^n-1)
= (1/6)n(n+1)(61-4n) +(1/2)(3^n-1)
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