Counter
Mathematics Dictionary
Dr. K. G. Shih
Sum[n^2] = (n*(n+1)*(2*n + 1))/6


  • Q01 | - Method 1 : Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
  • Q02 | - Method 2 : Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
  • Q03 | - Method 3 : Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
  • Q04 | - Method 4 : Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
  • Q05 | - Pattern


Q01. Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6

Method 1 : By observation
  • 1^2 + 2^2 = 1 + 4 = 5 = 2*3*5/6
  • 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 = 3*4*7/6
  • 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 = 4*5*9/6
  • Since 4*5*9/6 = 4*(4+1)*(2*4+1)/6 for n = 4
  • Proof complete
  • But this method is not ensured for large n

Go to Begin

Q02. Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6

Method 2 : Use Sum[C(n+1,2)] = C(n+3,2)
  • Binonial expansion coefficients
    • C(n+1,2) = (n+1)*n/2!
    • C(n+2,3) = (n+2)*(n+1)*n/3!
  • Sum[(n*(n+1))/2!] = (n*(n+1)*(n+2)/(3!)
  • Hence Sum[n^2 + n] = 2*(n+2)*(n+1)*n/6
  • Hence Sum[n^2] = 2*(n+2)*(n+1)*n/6 - Sum[n]
  • Since Sum[n] = n*(n+1)/2
  • Hence Sum[n^2] = n*(n+1)*[2*(n+2)/6 - 1/2]
  • Hence Sum[n^2] = n*(n+1)*(2*n+1)/6

Go to Begin

Q03 Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6

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*(k+1)+1)/6 exists
  • n = k+1 and S(k+1) = S(k) + (k+1)^2
  • Hence S(k+1) = k*(k+1)*(2*k+1)/6 + (k+1)^2
  • Hence S(k+1) = (k+1)*(k*(2*k+1)/6 + (k+1))
  • Hence S(k+1) = ((k+1)*(2*k^2+k)/6 + k+1))
  • Hence S(k+1) = (k+1)*(2*k^2 + 7*k + 6)/6
  • Hence S(k+1) = (k+1)*(k+1+1)*(2*(k+1)+1)/6
  • Proof complete

Go to Begin

Q04 Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6

Method 4 : Use (a+b)^3 = a^3 + 3*(a^2)*b + 3*a*(b^2) + 1 ............(1)
  • Since (x+1)^3 = x^3 + 3*x^2 + 3*x + 1
  • Sum[(x + 1)^3 - x^3] = Sum[3*x^2] + Sum[3*x] + Sum[1]
  • Sum[(x + 1)^3 - x^3]
    • = (2^3 - 1) + (3^3 - 2^3) + ...... ((x+1)^3 - x^3)
    • = (x + 1)^3 - 1 (All terms cancelled out expcept -1 and (x+1)^3)
    • = x^3 + 3*x^2 + 3*x
  • Hence Sum[3*x^2] + Sum[3*x] + Sum[1] = x^3 + 3*x^2 + 3*x ...... (2)
  • From (1) and (2) we have
  • Sum[3*x^2] + Sum[3*x] + Sum[1] = x^3 + 3*x^2 + 3*x
  • Sum[3*x^2] = x^3 + 3*x^2 + 3*x - Sum[3*x] - Sum[1]
  • Sum[3*n^2] = n^3 + 3*n^2 + 3*n - 3*n*(n+1)/2 - n
    • = n^3 + 3*n^2 + 3*n - 3*n*(n+1)/2 - n
    • = n^3 + 2*n + 3*n^2 - 3*n^2/2 - 3*n/2
    • = n^3 + 3*n^2/2 + n/2
    • = n*(2*n^2 + 3*n + 1)/2
    • = n*(n+1)*(2*n+1)/2
  • Hence Sum[n^2] = n*(n+1)*(2*n+1)/6

Go to Begin

Q05 Patterns

Squares in squares Pattern : Numbers in square patterns

    * .......... 1^2 = 1

    * *
    * * ........ 2^2 = 4

    * * *
    * * *
    * * * ...... 3^2 = 9

Go to Begin
Show Room of MD2002 Contact Dr. Shih Math Examples Room
Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
1