Mathemtics Dictionary Example
Series : Sum[n^3] = (n*(n+1)/2)^2
Subjects

Answers


Q01. Outline
    1. T(n) is the nth term
    2. S(n) is the sum of n terms = Sum[x^3] from 1 to n
    3. F(n) = ist difference = T(n+1)-T(n)
    4. G(n) = 2nd difference = F(n+1)-F(n)
    5. H(n) = 3rd difference = G(n+1)-G(n) = common difference

    6. Prove that S(n) = (n*(n+1))^2 where ^2 = 2 to power 2
      a. Method 1 : By observation
      b. Method 2 : Use Sum[a^3+3*a^2+3*a+1]
      c. Method 3 : Use Sum[C(n+2,3)]=C(n+3,4)
      d. Method 4 : Mathematical induction
    7. Cubes in cubes pattern (Program 12 24)
    8. Note : C(n,r) is the coefficients of binomail expansion

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Q02. Example : Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2
* Method 1 : By observation
    1^3 + 2^3 = 1 + 8 = 9 = (1+2)^2
    1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36 = (1+2+3)^2
    1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100 = (1+2+3+4)^2

    Since 1 + 2 + 3 + 4 + ..... + n = n*(n+1)/2
    Proof complete
* Method 2 : Use Sum[C(n+2,3)]=C(n+4,4)
    C(n+2,3) = (n+2)*(n+1)*n/3!
    C(n+3,4) = (n+3)*(n+2)*(n+1)/4!
    Sum[(n+2)*(n+1)*n/3!] = (n+3)*(n+2)*(n+1)/4!
    Expnand :

    Sum[n^3+3*n^2+2*n] = 6*(n+3)*(n+2)*(n+1)*n/24
    Hence Sum[n^3] = 6*(n+3)*(n+2)*(n+1)*n/24-3*Sum[n^2]-2*Sum[n]
    Since Sum[n^2] = n*(n+1)*(2*n+1)/6 and Sum[n] = n*(n+1)/2
    Hence Sum[n^3] = n*(n+1)*[(n+3)*(n+2)/4 - (2*n-1)/2 -1]

    Simplify right hand side and proof complete
* Method 3 : Use mathematical induction
    n = 1 the sum is true
    n = 2 the sum is true
    n = k+1 should be true. i.e. S(k+1)=((k+1)*k+2)/2)^2 exists
    n = k+1 S(k+1) = S(k)+(k+1)^3

    Hence S(k+1) = (k*(k+1)/2)^2+(k+1)^3 = ((k+1)^2)*(k^2+4*(k+1))/4
    We have S(k+1) = ((k+1)*(k+2)/2)^2 Proof complete
* Method 4 : Use (a+b)^4 = a^4+4*(a^3)*b+6*(a^2)*(b^2)+4*a*b^3 + 1
  • (x + 1)^4 = x^3 + 4*(x^3) + 6*(x^2) + 4*x + 1
  • Sum[(x + 1)^4 - x^4] = Sum[4*(x^3) + 6*(x^2) + 4*x + 1] .......... (1)
  • Sum[(x + 1)^4 - x^4] =
    • = (2^4 - 1) + (3^4 - 2^4) + .... + ((x+1)^4 - x^4)
    • = (x + 1)^4 - 1
    • = x^4 + 4*x^3 + 6*x^2 + 4*x ................................... (2)
    • Note : since all terms cancelled out except -1 and (x + 1)^4
  • From (1) and (2), we have
    • Sum[4*(x^3) + 6*(x^2) + 4*x + 1] = x^4 + 4*x^3 + 6*x^2 + 4*x
    • Sum[4*x^3] = x^4 + 4*x^3 + 6*x^2 + 4*x - Sum[6*x^2] - Sum[4*x] - Sum[1]
    • Sum[4*n^3] = n^4 + 4*n^3 + 6*n^2 + 4*n - Sum[6*n^2] - Sum[4*n] - Sum[1]
    • Sum[4*n^3] = n^4 + 4*n^3 + 6*n^2 + 4*n - n*(n+1)*(2*n+1) - 2*n*(n+1) - n
    • Sum[4*n^3] = n^4 + 4*n^3 + 6*n^2 + 4*n - n*(n+1)*(2*n+3) - n
    • Sum[4*n^3] = n^4 + 4*n^3 + 6*n^2 + 4*n - (2*n^3 + 5*n^2 + 3*n) - n
    • Sum[4*n^3] = n^4 + 2*n^3 + n^2
    • Sum[n^3] = (n*(n+1)/2)^2

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Q03. Pattern Cubes in cubes 1 + 8 + 27 + 64 + ....

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Q04. Find 3rd difference of 1,8,27,64,125,....
    Sequence : 01 08 27 64 125 ....
    1st diff : .. 07 19 37 061 ....
    2nd diff : .. .. 12 18 024 ....
    3rd diff : .. .. .. 06 006 ....

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Q05. Q4. Home work :
    a. Write all formulae used in above methods to your note book.
    b. Try to remember them
    c. Study more series which are given in Chapter 14 of MD2002

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Q06. Exercises :

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Q07. Reference
  • MD2002 ZM14.exe
  • MD2002 ZS.txt

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Q08. Formula
Sum from 1 to n
  • 1. Sum[1] = n
  • 2. Sum[n] = n*(n+1)/2
  • 3. Sum[n^2] = n*(n+1)*(2*n+1)/6
  • 4. Sum[n^3] = (n*(n+1)/2)^2
  • 5. Sum[n^4] = n*(n+1)*(6*n^3+9*n^2+n-1)/30
Special series
  • (1-1/4)*(1-1/9)*(1-1/16)*.....*(1-1/n^2) = (n+1)/(2*n)
  • 1/(1*2*3) + 1/(2*3*4) + .... + 1/(n*(n+1)*(n+2)) = [1 - 2/((n+1)*(n+2))]/4

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Q09. Answer

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Q10. Answer

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