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International Baccalaureatte | Further Maths HL
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Anyone out there knows hw to solve qus 1 and 4, thxs
Date Posted: 6 years ago
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Sn = a(rⁿ -1)/(r - 1)
S2n = a(r²ⁿ - 1)/(r - 1)
S2n/Sn = a(r²ⁿ -1)/(r - 1) ÷ a(rⁿ - 1)/(r - 1) = K
a(r²ⁿ -1)/(r - 1) x (r - 1) / (a(rⁿ - 1)) = K
(r²ⁿ - 1) / (rⁿ - 1) = K
(rⁿ + 1)(rⁿ - 1) / (rⁿ - 1) = K
rⁿ + 1 = K (proved)
S = a / (1 - r)
2S = a² / (1 - r²)
25/8 = a + ar = a(1 + r) -----①
2S / S = a² / (1 - r²) ÷ a / (1 - r)
2 = a² / (1 - r²) x (1 - r) / a
2 = a/(1 + r)
a = 2(1 + r) ----②
Sub ② into ①,
25/8 = 2(1 + r)(1 + r) = 2r² + 4r + 2
2r² + 4r - 9/8 = 0
r² + 2r - 9/16 = 0
(r - ¼)(r + 9/4) = 0
r = ¼ or r = -9/4 (N.A, ∣ r ∣ < 1 for series to be convergent)
so r = ¼.
so a = 2(1 + ¼) = 5/2
S = a / (1 - r) = 5/2 / (1 - ¼)
S = 10/3 = 3⅓
S2n = a(r²ⁿ - 1)/(r - 1)
S2n/Sn = a(r²ⁿ -1)/(r - 1) ÷ a(rⁿ - 1)/(r - 1) = K
a(r²ⁿ -1)/(r - 1) x (r - 1) / (a(rⁿ - 1)) = K
(r²ⁿ - 1) / (rⁿ - 1) = K
(rⁿ + 1)(rⁿ - 1) / (rⁿ - 1) = K
rⁿ + 1 = K (proved)
S = a / (1 - r)
2S = a² / (1 - r²)
25/8 = a + ar = a(1 + r) -----①
2S / S = a² / (1 - r²) ÷ a / (1 - r)
2 = a² / (1 - r²) x (1 - r) / a
2 = a/(1 + r)
a = 2(1 + r) ----②
Sub ② into ①,
25/8 = 2(1 + r)(1 + r) = 2r² + 4r + 2
2r² + 4r - 9/8 = 0
r² + 2r - 9/16 = 0
(r - ¼)(r + 9/4) = 0
r = ¼ or r = -9/4 (N.A, ∣ r ∣ < 1 for series to be convergent)
so r = ¼.
so a = 2(1 + ¼) = 5/2
S = a / (1 - r) = 5/2 / (1 - ¼)
S = 10/3 = 3⅓
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