Mathematics Dictionary Dr. K. G. Shih

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Sequences

Symbol Defintion Example : Sqr(x) = square root of x

Q01 | - Sequences : 1, 3, 7, 13, ....Proof of sine law Q02 | - Sequences : 1, 4, 10, 20, ..... Q03 | - Sequences Q04 | - Sequences Q05 | - Sequences

Q01. Seqences : 1, 3, 7, 13, 21, .... Questions 1. Find 1st and 2nd difference 2. Find y = F(x) 3. Find T(n) 4. Find S(n) Solution : Find difference n = .......1, 2, 3, 04, 05, ... Squences : 1, 3, 7, 13, 21, ... 1st diff : .. 2, 4, 06, 08, ... 2nd diff : ..... 2, 02, 02, ... Hence expression is y = a*x^2 + b*x + c and a = 1 Find y = a*x^2 + b*x + c Since 2nd diff has common difference 2, hence a = 1 If x = 1, y = 1 Hence 1 = 1^2 + b*1 + c Hence b + c = 0 ................................. (1) If x = 2, y = 3 hence 3 = 2^2 + b*2 + c Hence 2*b + c = -1 .............................. (2) Solve (1) and (2), we have b = -1 and c = 1 Hence y = x^2 - x + 1 Find T(n) and S(n) T(n) = n^2 - n + 1 S(n) = Sum[n^2 - n + 1] = Sum[n^2] - Sum[n] + Sum[1] = n*(n+1)*(2*n+1)/6 - n*(n+1)/2 + n = n*((n+1)*(2*n+1)/6 - (n+1)/2 + 1) = n*(2*n^2 +3*n + 1)/6 - (3*n + 3)/6 + 6/6) = n*(2*n^2 + 4)/6 = n*(n^2 + 2)/3 Verify S(n) S(4) = 4*(4^2 + 2)/3 = 4*18/3 = 24 S(4) = 1 + 3 + 7 + 23 = 24 Go to Begin Q02. Sequences : 1, 4, 10, 20, ..... Questions 1. Find difference 2. Find y = F(x) 3. Find T(n) 4. Find S(n) Solution : Find difference n = .......1, 2, 03, 04, 05, 06, ... Squences : 1, 4, 10, 20, 35, 50, ... 1st diff : .. 3, 06, 10, 15, 21, ... 2nd diff : ..... 03, 04, 05, 06, ... 3rd diff : ......... 01, 01, 01, ... Hence expression is y = a*x^3 + b*x^2 + c*x + d and a = 1/6 Find y = a*x^3 + b*x^2 + c*x + d Since 3nd diff has common difference 1, hence a = 1/6 If x = 1, y = 1 Hence 1 = (1^3)/6 + b*1^2 + c*1 + d Hence b + c + d = 5/6 ................................. (1) If x = 2, y = 4 hence 4 = (2^3)/6 + b*2^2 + c*2 + d Hence 4*b + 2*c + d = 8/3 ............................. (2) If x = 3, y = 10 Hence 10 = (3^3)/6 + b*3^2 + c*3 + d Hence 9*b + 3*c + d = 11/2 ............................ (3) (2) - (1), we have 3*b + c = 11/6 ........................................ (4) (3) - (2), we have 5*b + c = 17/6 ........................................ (5) (5) - (4), we have 2*b = 1 and b = 1/2 Substitute b into (4), we have 3*(1/2) + c = 11/6. c = 1/3 Substitute b = 1/2 and c = 1/3 into (1), we have 1/2 + 1/3 + d = 5/6 and d = 0 Hence y = (x^3)/6 + (x^2)/2 + x/3 = (x^3 + 3*x^2 + 2*x)/6 Find T(n) and S(n) T(n) = (n^3 + 3*n^2 + 2*n)/6 = n*(n+1)*(n+2)/6 S(n) = Sum[n*(n+1)*(n+2)/6] = n*(n+1)*(n+2)*(n+3)/24 Verify S(n) S(3) = 3*4*5*6/24 = 15 S(3) = 1 + 4 + 10 = 15 Proof of S(n) See keyword Pascal and sequences Go to Begin Q03. Go to Begin Q04. Go to Begin Q05. Go to Begin

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