等差級數問題*3

1.
第 16 項 a16
= S16 - S15
= (16² + 2 x 16) - (15² + 2 x 15)
= 33


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2.
首項 a1
= 前 1 項和
= 3 x 1² + 4 x 1
= 7

首 2 項和：
[2 x 7 + (2 - 1)d] x 2 / 2 = 3 x 2² + 4 x 2
14 + d = 20
d = 6
公差 = 6

第 15 項
= 7 + 14 x 6
= 91


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3.
設等差級數為其中連續三項為 a - d, a,a + d，其公差為 d。

則 ..... 1/(a- d), 1/a, 1/(a + d) ...... 亦為等差級數，其公差
1/a - 1/(a - d) = 1/(a + d) - 1/a
2/a = 1/(a + d) + 1/(a - d)
2/a = [(a + d) + (a - d)] / (a + d)(a - d)
2/a = 2a / (a + d)(a - d)
a² = (a + d)(a - d)
a² = a² - d²
d² = 0
d = 0
公差 = 0

a16=S16-S15

S16=16²+2×16=288
S15=15²+2×15=255
→
S16=S15=a16=33

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a2=a1+1d

a1=S1=3×1²+4×1=7
a2=S2-S1=3×2²+4×2-(7)=20-7=13
→
13=7+d
→d=6

a15=a1+14d=7+14×6=91

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等於0