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Mathematics Dictionary
Dr. K. G. Shih
Sequence in Pascal Triangle
Questions

  • Q01 | - Defintion of Pascal triangle
  • Q02 | - Sequence along r = 0 : 1, 1, 01, 01, 01, ......
  • Q03 | - Sequence along r = 1 : 1, 2, 03, 04, 05, ......
  • Q04 | - Sequence along r = 2 : 1, 3, 06, 10, 15, ......
  • Q05 | - Sequence along r = 3 : 1, 4, 10, 20, 35, ......
  • Q06 | - Sequence along r = 4 : 1, 5, 15, 35, 70, .....
  • Q07 | - Sequence along r = 5 : 1, 6, 21, 56, 126 ....
  • Q08 | - Find Sum[n^4] usimg Sum[C(n+3,4)] = C(n+4,5)
  • Q09 | - Fibonacci's sequence in Pascal triangle
  • Q10 | - Formula
Answers


Q1. Deifintion of Pascal Triangle
A1. Answer
[Defintion] Pascal triangle : It is coeff of binomial expansion
  • 1, 2, 1 are ceeficients of expansion of (x+y)^2
  • 1, 3, 3, 1 are ceeficients of expansion of (x+y)^3
  • 1, 4, 6, 4, 1 are ceeficients of expansion of (x+y)^4
[Pscal triangle]
    Column ...... r0 r1 r2 r3 r4 r5 r6 r7 r8
    ---------------------------------
    n=0 ......... 01
    n=1 ......... 01 01
    n=2 ......... 01 02 01
    n=3 ......... 01 03 03 01
    n=4 ......... 01 04 06 04 01
    n=5 ......... 01 05 10 10 05 01
    n=6 ......... 01 06 15 20 15 06 01
    n=7 ......... 01 07 21 35 35 21 07 01
    n=8 ......... 01 08 28 56 70 56 28 08 01
[Sequences]
  • Sequences along r0 is 1, 1, 1, 1, 1, ......
  • Sequences along r1 is 1, 2, 3, 4, 5, ......
  • Sequences along r2 is 1, 3, 6, 10, 15, ......
[Relation witn C(n,r)]
 Go to Begin

Q2. What is r0 series ?
A2. Answer
  • Sequence along r0 column : 1, 1, 1, 1, 1, 1, 1, 1, ....
  • The common differece is 0
  • T(n) = nth term = 1
  • S(n) = Sum[1] = n
Go to Begin

Q3. What is r1 series ?
A3. Answer
  • Sequence along r1 column : 1, 2, 3, 4, 5, 6, 7, 8, ....
  • This is an arithmetic sequence with common differece is 1
  • T(n) = nth term = n
  • S(n) = Sum[n] = n*(n+1)/2
  • 1st difference is common difference d = 1
Go to Begin

Q4. What is r2 series ?
A4. Answer
 [Example] Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3!
  • Sum[n*(n+1)/2]
  • = Sum[n^2/2]+Sum[n/2]
  • = (Sum(n^2) + Sum[n])/2
  • = (n*(n+1)*(2*n+1)/6 + n*(n+1)/2)/2
  • = n*(n+1)*(2*n+1)/12 + n*(n+1)/4
  • = n*(n+1)*(2*n+1)/3+1)/4
  • = n*(n+1)*(2*n+4)/12
  • = n*(n+1)*(n+2)/3! where 3!=1*2*3=6
  • = C(n+2, 3) the coefficients of binomial expansion
Go to Begin

Q5. What is r3 series ?
A5. Answer
  • Squence along r3 column : 1, 4, 10, 20, 35, 56, ....
  • T(n) = nth term = n*(n+1)*(n+2)/3!
  • S(n) = Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Hence Sum[C(n+2, 3)] = C(n+3, 4)
  • 1st difference is r2 series
  • 2nd difference is r1 series
  • 3nd difference is common difference d=1
[Example] Prove that Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Sum[n*(n+1)*(n+2)/6]
  • = Sum[n^3/6]+Sum[3*n^2/6]+Sum[2*n/6]
  • = (n*(n+1)/2)/6 + n*(n+1)*(2*n+1)/12 + n*(n+1)/12
  • = n*(n+1)*[(n*(n+1)/24 + (2*n+1)/12 +1/12]
  • = n*(n+1)*(n^2+5*n+6)/24
  • = n*(n+1)*(n+2)*(n+3)/4! where 4!=1*2*3*4
Go to Begin

Q6. What is r4 series ?
A6. Answer
  • Sequence along r4 column : 1, 5, 15, 35, 70, ....
  • T(n) = nth term = n*(n+1)*(n+2)*(n+3)/4!
  • S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
  • Hence Sum[C(n+3, 4)] = C(n+4, 5)
  • 1st difference is r3 series
  • 2nd difference is r2 series
  • 3rd difference is r1 series
  • 4th difference is common difference d=1
Go to Begin

Q7. What is r5 series ?
A7. Answer
[Definition]
  • Sequence along r5 column
  • T(n) = nth term = n*(n+1)*(n+2)*(n+3)(n+4)/5!
  • S(n) = Sum[T(n)] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
  • Hence Sum[C(n+4, 5)] = C(n+5, 6)
  • 1st difference is r4 series
  • 2nd difference is r3 series
  • 3rd difference is r2 series
  • 4th difference is r1 series
  • 5th difference is common difference d=1
[Coclusion]
  • Sum[C(n+1,2)] = C(n+2,3)
  • Sum[C(n+2,3)] = C(n+3,4)
  • Sum[C(n+3,4)] = C(n+4,5)
  • Sum[C(n+4,5)] = C(n+5,6)
Go to Begin

Q8. Find Sum[n^4] by using Sum[C(n+3,4)] = C(n+4,5)
A8. Answer
  • Sum[C(n+3),4] = Sum[n*(n+1)*(n+2)*(n+3)/4!] = C(n+4,5)
  • (Sum[n^4] + 6*Sum[n^3] + 11*Sum[n^2] + 6*Sum[n])/24 = C(n+4,5)
  • If we know Sum[n^3], Sum[n^2], Sum[n^1], then we can find Sum[n^4]
  • The answer is Sum[n^4] = (6*n^5 + 5*n^4 + 10*n^3 - n)/30
Go to Begin

Q9. Fibonacci's sequence in Pascal triangle
A5. Answer in Fibonacc's pattern
  • Fibonacci's Sequence : 1, 1, 2, 3, 5, 8, 13, ...
  • Recursion formula : T(0) = 0, T(1) = 1 and T(n+2) = T(n+1) + T(n)
Go to Begin

Q10. Formula

[Sum in C(n,r) Form]
  • Sum[C(n+1,2)] = C(n+2,3)
  • Sum[C(n+2,3)] = C(n+3,4)
  • Sum[C(n+3,4)] = C(n+4,5)
  • Sum[C(n+4,5)] = C(n+5,6)
[Sum in factor Form]
  • Sum[1] = n
  • Sum[n] = n*(n+1)/2!
  • Sum[n*(n+1)/2!] = n*(n+1)*(n+2)/3!
  • Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
  • Sum[n*(n+1)*(n+2)*(n+3)/4!] = n*(n+1)*(n+2)*(n+3)*(n+4)/5!
[Other series] Go to Begin
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