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Mathematics Dictionary
Dr. K. G. Shih
Sum[n^3] = (n*(n+1)/2)62


  • Q01 | - Method 1 : Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2
  • Q02 | - Method 2 : Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2
  • Q03 | - Method 3 : Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2
  • Q04 | - Method 4 : Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2
  • Q05 | - Pattern


Q01. Prove that 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

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Q02. Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2

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

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Q03 Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2

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

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Q04 Prove that Prove that 1^3 + 2^3 + 3^3 +...+ n^3 = (n*(n+1)/2)^2

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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Q05 Patterns


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