①等差数列的前n项和为Sn，且a2=4,s2=6,则(Sn+64)/an的最小值是?衷心的感谢你的解答，过程详细 ② 5
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an= a1+(n-1)d
a2=4
a1+d=4 (1)
S2=6
2a1+d=6 (2)
(2)-(1) => a1=2
from (1), d=2
an = 2+2(n-1) = 2n
(Sn +64)/an
=[(n+1)n+64]/(2n)
=(n^2+n+64)/(2n)
consider
f(x) = (x^2+x+64)/(2x)
= (1/2)(x+ 2)+ 32/x
f'(x) =1/2 - 32/x^2
f'(x) =0
1/2 - 32/x^2 =0
x^2-64=0
x=8
f''(x) = 64/x^3 >0 (min)
min (Sn +64)/an at n=8
=(64+8+64)/(8)
=17
a2=4
a1+d=4 (1)
S2=6
2a1+d=6 (2)
(2)-(1) => a1=2
from (1), d=2
an = 2+2(n-1) = 2n
(Sn +64)/an
=[(n+1)n+64]/(2n)
=(n^2+n+64)/(2n)
consider
f(x) = (x^2+x+64)/(2x)
= (1/2)(x+ 2)+ 32/x
f'(x) =1/2 - 32/x^2
f'(x) =0
1/2 - 32/x^2 =0
x^2-64=0
x=8
f''(x) = 64/x^3 >0 (min)
min (Sn +64)/an at n=8
=(64+8+64)/(8)
=17
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