add math exam(today)...last question of mi (10marks)

Let P ( n ) be the proposition “1^4 + 2^4 + 3^4 + ... + n^4 = 1/30 n(n + 1)(6n^3 + 9n^2 + n - 1)
”.
When n = 1,
L.H.S. = 1^4 = 1
R.H.S. = (1/30)(1)(2)(15) = 1 = L.H.S.
So P ( 1 ) is true.
Assume P ( k ) is true for some positive integers k, i.e.
1^4 + 2^4 + 3^4 + ... + k^4 = 1/30 k(k + 1)(6k^3 + 9k^2 + k - 1)

When n = k + 1,
L.H.S. = 1^4 + 2^4 + 3^4 + ... + k^4 + (k+1)^4
= (1/30) k(k + 1)(6k^3 + 9k^2 + k - 1) + (k+1)^4
= (1/30)(k+1)[6k^4+9k^3+k^2-k+30k^3+90k^2+90k+30].....通分+散開埋
= (1/30)(k+1)[6k^4+39k^3+91k^2+89k+30]
=(1/30)(k+1)(k+2)(6k^3+27k^2+37k+15)
=(1/30)(k+1)(k+2)[6(k^3+3k^2+3k+1)+9k^2+19k+9]
=(1/30)(k+1)(k+2)[6(k+1)^3+9(k^2+2k+1)+k]
=(1/30)(k+1)(k+2)[6(k+1)^3+9(k+1)^2+(k+1)-1]
R.H.S. =(1/30)(k+1)(k+2)[6(k+1)^3+9(k+1)^2+(k+1)-1]


= L.H.S. So P ( k + 1 ) is true.
By the principle of mathematical induction, P ( n ) is true for all positive integers n.