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Mathematics Dictionary
Dr. K. G. Shih
Sum[n^4] = ?


  • Q01 | - Method 1 : S(n) = 1^4 + 2^4 + 3^4 + ... + n^4
  • Q02 | - Method 2 : S(n) = 1^4 + 2^4 + 3^4 + ... + n^4
  • Q03 | - Method 3 : S(n) = 1^4 + 2^4 + 3^4 + ... + n^4
  • Q04 | - Method 4 : S(n) = 1^4 + 2^4 + 3^4 + ... + n^4
  • Q05 | - Reference


Q01. S(n) = 1^4 + 2^4 + 3^4 + ... + n^4

Method 1 : By observation
  • 1^4 + 2^4 = 1 + 16 = 17
  • 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98
  • 1^4 + 2^4 + 3^4 + 4^4 = 98 + 256 = 354
  • It is not easy to make conclusion

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Q02. S(n) = 1^4 + 2^4 + 3^4 +...+ n^4

Method 2 : Use Sum[C(n+3,4)] = C(n+4,5)
  • Sum[C(n+3),4] = Sum[n*(n+1)*(n+2)*(n+3)/4!]
  • Since n*(n+1)*(n+2)*(n+3) = n^4 + 6*n^3 + 11*n^2 + 6*n
  • Sum[C(n+3),4] = (Sum[n^4] + 6*Sum[n^3] + 11*Sum[n^2] + 6*Sum[n])/24 ...... (1)
  • C(n+4,5) = n*(n+1)*(n+2)*(n+3)*(n+4)/(5!)
  • Since n*(n+1)*(n+2)*(n+3)*(n+4) = n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n .. (2)
  • Sum[C(n+3),4]
    • = C(n+4,5)
    • = n*(n+1)*(n+2)*(n+3)*(n+4)/(5!)
    • = (Sum[n^4] + 6*Sum[n^3] + 11*Sum[n^2] + 6*Sum[n])/24 ................ (3)
  • Sum[n^4] = 24*n*(n+1)*(n+2)*(n+3)*(n+4)/120
  • - 6*Sum[n^3] - 11*sum[n^2] - 6*Sum[n]
  • From (2) and (3)
  • Sum[n^4]
    • = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n)/5
    • - 6*Sum[n^3] - 11*sum[n^2] - 6*Sum[n]
  • We know that
    • Sum[1] = n.
    • Sum[n] = n*(n+1)/2.
    • Sum[n^2] = n*(n+1)*(2*n+1)/6.
    • Sum[n^3] = (n*(n+1)/2)^2
  • Sum[n^4]
    • = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n)/5
    • - 6*(n*(n+1)/2)^2 - 11*n*(n+1)*(2*n+1)/6 - 6*n*(n+1)/2
    • = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n)/5
    • - 6*(n^4 +2*n^3 + n^2)/4 - 11*(2*n^3 + 3*n^2 + n)/6 - 3*(n^2 + n)
    • = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n)/5
    • - (6*n^4 + 12*n^3 + 6*n^2)/4 - (22*n^3 + 33*n^2 + 11*n)/6 - 3*(n^2 + n)
    • = 24*(n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n)/120
    • - 30*(6*n^4 + 12*n^3 + 6*n^2)/120 - 20*(22*n^3 + 33*n^2 + 11*n)/120
    • - 360*(n^2 + n)/120
    • = (24*n^5)/120 + (24*10 - 30*6)*(n^4)/120
    • + (24*35 - 30*12 - 20*22)*(n^3)/120 + (24*50 - 30*6 - 20*33 - 360)*(n^2)/120
    • + (24*24 - 220 - 360)*n/120
    • = (24*n^5 + 60*n^4 + 40*n^3 + - 4*n)/120
    • = n*(6*n^4 + 15*n^3 + 10*n^2 - 1)/30

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Q03 S(n) = 1^4 + 2^4 + 3^4 +...+ n^4 = n*(6*n^4 + 15*n^3 + 10*n^2 - 1)/30

Method 3 : Use mathematical induction
  • n = 1
    • S(1) = 1^4 = 1
    • S(1) = n*(6*n^4 + 15*n^3 + 10*n^2 - 1)/30 = 1
    • It is is true
  • n = 2
    • S(1) = 1^4 + 2^4 = 1 + 16 = 17
    • S(1) = 2*(6*2^4 + 15*2^3 + 10*2^2 - 1)/30 = 1
    • S(1) = 2*(96 + 120 + 40 - 1)/30 = 2*255/30 = 17
    • It is is true
  • n = k+1 should be true
  • n = k+1
    • LHS = S(k) + (k + 1)^4
    • RHS = 6*k^5 + 15*k^4 + 10 k^3 - k)/30 + (k + 1)^4
    • = ....
  • Proof complete

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Q04 S(n) = 1^4 + 2^4 + 3^4 +...+ n^4

Method 4 : Use (a+1)^5 = a^5 + 5*(a^4) + 10*(a^3) + 10*a^2 + 5*a + 1
  • (x + 1)^5 = x^5 + 5*(x^4) + 10*(x^3) + 10*(x^2) + 5*x + 1
  • Sum[(x + 1)^5 - x^5] = Sum[5*(x^4) + 10*(x^3) + 10*(x^2) + 5*x + 1] .... (1)
  • Sum[(x + 1)^5 - x^5] =
    • = (2^5 - 1) + (3^5 - 2^5) + .... + ((x+1)^5 - x^5)
    • = (x + 1)^5 - 1
    • = x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x ............................... (2)
    • Note : since all terms cancelled out except -1 and (x + 1)^5
  • From (1) and (2), we have

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Q05 Reference


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