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junior college 1 | H2 Maths
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Hello! :) how do i solve part ii and iii? (sorry for asking so late at night!)
Date Posted: 3 years ago
Views: 283
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i)
Given : u1 = 1, u2 = 7, u(n+1) = 3un + Pn
u2 = 3u1 + P(1) → 7 = 3(1) + P
P = 7 - 3 = 4
u3 = 3u2 + P(2) = 3(7) + 2(4) = 21 + 8 = 29
ii)
Given : un = a(3ⁿ) + bn + c
So,
u1 = a(3¹) + b(1) + c → 1 = 3a + b + c ①
u2 = a(3²) + b(2) + c → 7 = 9a + 2b + c ②
u3 = a(3³) + b(3) + c → 29 = 27a + 3b + c ③
There are now 3 equations with 3 unknowns. You can solve this using GC and still get the full credit/marks, but manually it would be something like :
③ - ② : 22 = 18a + b→ ④
② - ① : 6 = 6a + b →⑤
④ - ⑤ : 16 = 12a → a = 16/12 = 4/3
Sub a = 4/3 into ⑤,
6 = 6(4/3) + b → b = 6 - 8 = -2
Sub a = 4/3, b = -2 into ①,
1 = 3(4/3) - 2 + c → c = 1 - 4 + 2 = -1
∴ a = 4/3, b = -2, c = -1
Next post for part (iii)
Given : u1 = 1, u2 = 7, u(n+1) = 3un + Pn
u2 = 3u1 + P(1) → 7 = 3(1) + P
P = 7 - 3 = 4
u3 = 3u2 + P(2) = 3(7) + 2(4) = 21 + 8 = 29
ii)
Given : un = a(3ⁿ) + bn + c
So,
u1 = a(3¹) + b(1) + c → 1 = 3a + b + c ①
u2 = a(3²) + b(2) + c → 7 = 9a + 2b + c ②
u3 = a(3³) + b(3) + c → 29 = 27a + 3b + c ③
There are now 3 equations with 3 unknowns. You can solve this using GC and still get the full credit/marks, but manually it would be something like :
③ - ② : 22 = 18a + b→ ④
② - ① : 6 = 6a + b →⑤
④ - ⑤ : 16 = 12a → a = 16/12 = 4/3
Sub a = 4/3 into ⑤,
6 = 6(4/3) + b → b = 6 - 8 = -2
Sub a = 4/3, b = -2 into ①,
1 = 3(4/3) - 2 + c → c = 1 - 4 + 2 = -1
∴ a = 4/3, b = -2, c = -1
Next post for part (iii)
(iii)
From previous part we know that Un = 4/3 (3ⁿ) - 2n - 1 So likewise, Ur = 4/3 (3^r) - 2r - 1
Starting from r = 1, for a total of n terms,
∑ Ur
= ∑ ( 4/3 (3^r) - 2r - 1)
= ∑ 4/3 (3^r) - ∑ 2r - ∑ 1
= 4/3 ∑ (3^r) - 2∑ r - ∑ 1
Now, 3^r is our usual geometric progression (G.P), with a common ratio r = 3 and first term a = 3¹ = 3
So we can use our sum of GP formula Sn = a(rⁿ - 1)/(r - 1) = 3(3ⁿ - 1)/(3 - 1) = 3(3ⁿ - 1) / 2
∑r is the sum of n consecutive integers, from 1 to n. So we can use the formula ½n(n+1) (this is often called the pairing or rainbow method at O levels/PSLE)
∑1 is simply adding up n '1's together, so it just equals n.
Putting them altogether,
∑ Ur = 4/3 ∑ (3^r) - 2∑r - ∑ 1
= 4/3 (3(3ⁿ - 1)/2) - 2(½n(n+1)) - n
= 2(3ⁿ - 1) - n(n+1) - n
= 2(3ⁿ) - 2 - n² - n - n
= 2(3ⁿ) - n² - 2n - 2
From previous part we know that Un = 4/3 (3ⁿ) - 2n - 1 So likewise, Ur = 4/3 (3^r) - 2r - 1
Starting from r = 1, for a total of n terms,
∑ Ur
= ∑ ( 4/3 (3^r) - 2r - 1)
= ∑ 4/3 (3^r) - ∑ 2r - ∑ 1
= 4/3 ∑ (3^r) - 2∑ r - ∑ 1
Now, 3^r is our usual geometric progression (G.P), with a common ratio r = 3 and first term a = 3¹ = 3
So we can use our sum of GP formula Sn = a(rⁿ - 1)/(r - 1) = 3(3ⁿ - 1)/(3 - 1) = 3(3ⁿ - 1) / 2
∑r is the sum of n consecutive integers, from 1 to n. So we can use the formula ½n(n+1) (this is often called the pairing or rainbow method at O levels/PSLE)
∑1 is simply adding up n '1's together, so it just equals n.
Putting them altogether,
∑ Ur = 4/3 ∑ (3^r) - 2∑r - ∑ 1
= 4/3 (3(3ⁿ - 1)/2) - 2(½n(n+1)) - n
= 2(3ⁿ - 1) - n(n+1) - n
= 2(3ⁿ) - 2 - n² - n - n
= 2(3ⁿ) - n² - 2n - 2
OH i understand now thank you so much!! :)
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