数列{Bn}和函数F[x],已知F[x]=-3x+27,Bn=F[n],试判断{Bn}是否为等差数列,并求{Bn}的前n项和的最大值
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Bn=F[n] = -3n + 27
B - B = [-3(n+1) + 27] - [-3n + 27]
= -3 因此 Bn 是等差数列,公差为-3.
B1 = -3*1 + 27 = 24
Bn = -3n + 27
Sn = (B1 +Bn)*n/2
= (24 -3n + 27)*n/2
= 3(17-n)*n/2
= (3/2)*(-n^2 + 17n)
= (3/2)*[-n^2 + 17n - (17/2)^2 + (17/2)^2]
= (3/2)*[289/4 - (n- 17/2)^2]
因此当 n = 8 或 9 时,前n项和取最大值
此时S8 = S9 = (3/2)*[289/4 - 1/4] = (3/2)*72 = 108