数列题,高一数学,求解
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解:
(1)
设该等差数列的公差为d,S1,S2,S4的公比为q,则:
S1=a1
S2=a1+d
S4=a1+3d
因此:
(S2)²=S1·S4
(a1+d)² = a1·(a1+3d)
所以:
(a1)²+d²+2a1d = (a1)²+3a1d
d² = a1d
∴d = a1
q = S2/S1 = (a1+d)/a1 = 2
(2)
S2 = a1+d = 2a1=2d =4
因此:
a1=d=2
an = a1+(n-1)d=2n
(3)
bn = 3/[an·a(n+1)]
= 3/ [2n·2(n+1)]
=(3/4)·[(1/n) - 1/(n+1)]
因此:
Tn = (3/4)·[1-1/2+1/2 - 1/3+....+(1/n) - 1/(n+1) ]
= (3/4)· [1 - 1/(n+1)]
上式中,当n=1时,1/(n+1)有最大值,而Tn有最小值,因此:
Tn ≥ 3/8
若要Tn > m/20成立,只要:
3/8 > m/20
m < 15/2
因此:
m的最大正整数为:7
(1)
设该等差数列的公差为d,S1,S2,S4的公比为q,则:
S1=a1
S2=a1+d
S4=a1+3d
因此:
(S2)²=S1·S4
(a1+d)² = a1·(a1+3d)
所以:
(a1)²+d²+2a1d = (a1)²+3a1d
d² = a1d
∴d = a1
q = S2/S1 = (a1+d)/a1 = 2
(2)
S2 = a1+d = 2a1=2d =4
因此:
a1=d=2
an = a1+(n-1)d=2n
(3)
bn = 3/[an·a(n+1)]
= 3/ [2n·2(n+1)]
=(3/4)·[(1/n) - 1/(n+1)]
因此:
Tn = (3/4)·[1-1/2+1/2 - 1/3+....+(1/n) - 1/(n+1) ]
= (3/4)· [1 - 1/(n+1)]
上式中,当n=1时,1/(n+1)有最大值,而Tn有最小值,因此:
Tn ≥ 3/8
若要Tn > m/20成立,只要:
3/8 > m/20
m < 15/2
因此:
m的最大正整数为:7
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