Question

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【2】有一等差數列<an>,公差d,S,為此數列的前n项和,若
1725
G2024 = S2024 = 2024,則下列選項何者正確?
(1) 1012
> 0
(3)d>0 (4)a=-2024 (5) Sz0zz>0
(2) S1012 > S1011 (3) d>0 a₁ = -2024
S2022>0

【2】 已知某數列(an>,其前n项和Sn=n²-14n,
求a+a2+as+......
...
++ |α 30 | =
(20+(n-1)d)=n-149【興大附中
1-14 Ans: 57

Answer 1

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qn

(1)
S₂₀₂₄ = S₂₀₂₃ + a₂₀₂₄
⇒ S₂₀₂₃ = S₂₀₂₄ - a₂₀₂₄ = 0

a₁+a₂+⋯+a₂₀₂₂+a₂₀₂₃ = 0
(a₁+a₂₀₂₃) + (a₂+a₂₀₂₂) + ⋯ + (a₁₀₁₁+a₁₀₁₃) + a₁₀₁₂ = 0

a₁+a₂₀₂₃ = a₂+a₂₀₂₂ = ⋯ = a₁₀₁₁+a₁₀₁₃ = 2a₁₀₁₂
⇒ 2023a₁₀₁₂ = 0
⇒ a₁₀₁₂ = 0

(2)
S₁₀₁₂ = S₁₀₁₁ + a₁₀₁₂ = S₁₀₁₁

(3)
a₂₀₂₄ - a₁₀₁₂ = 1012d = 2024
⇒ d = 2

(4)
a₂₀₂₄ = a₁ + 2023d
⇒ a₁ = 2024 - 2023×2 = -2022

(5)
S₂₀₂₂ = S₂₀₂₃ - a₂₀₂₃
= 0 - (a₂₀₂₄-d)
= -2024 + 2
= -2022

Sₙ = n²-14n
⇒ { a₁ = S₁ = -13
{ aₙ = Sₙ - Sₙ₋₁ = (n²-14n) - [(n-1)²-14(n-1)]
= (n²-14n) - (n²-16n+15)
= 2n-15
(a₁=-13= 2·1-15)

令 aₙ = 2n-15 < 0
⇒ n ≤ 7

⇒ 所求 = -(a₁+a₂+⋯+a₇) + (a₈+a₉+⋯+a₃₀)
= - [2(-13)+6·2]·7/2 + [2(1)+22·2]·23/2
= 49 + 529
= 578

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