Maths

Since
[(n+1)^4 - n^4] + [n^4 - (n-1)^4] + [(n-1)^4 - (n-2)^4] +...+ 2^4] - [2^4 - 1^4 ] =

(n+1)^4 - 1^4

(4n^3 + 6n^2 + 4n + 1) + [4(n-1)^3 + 6(n-1)^2 + 4(n-1) + 1] + ...
+ 4(1)^3 + 6(1)^2 + 4(1) + 1 = (n+1)^4 - 1

4(1^3 + 2^3 +...+ n^3) + 6(1^2 + 2^2 + ... + n^2) + 4(1+2+...+n) + n
= (n+1)^4 - 1

4(1^3 + 2^3 + ... + n^3) + 6 n(n+1)(2n+1)/6 + 4 n(n+1)/2 + n
= (n+1)^4 - 1

4(1^3 + 2^3 + ... + n^3) = (n+1)^4 - 1 - n(n+1)(2n+1) - 2n(n+1) - n

4(1^3 + 2^3 + ... + n^3) = (n+1)[(n+1)^3 - n(2n+1) - 2n - 1)

4(1^3 + 2^3 + ... + n^3) = (n+1)(n^3 + 3n^2 + 3n + 1 - 2n^2 - n - 2n - 1)

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

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

1^3 + 2^3 + ... + n^3 = (1/4)[n(n+1)]^2

1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2

1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2




2010-01-15 19:55:06 補充：
Line 2 should be :

[(n+1)^4 - n^4] + [n^4 - (n-1)^4] + [(n-1)^4 - (n-2)^4] + ... +[3^4 - 2^4] + [2^4 - 1^4 ] =

I'm sorry that I should use "deduce the formula" instead of "prove that"...

(1 + 2 + 3 + 4 + .....+ n)^2 = = [n(n+1)/2]^2 = n^2(n + 1)^2/4.
Using the method of M.I.
For n = k + 1.
L.H.S = 1^3 + 2^3 = ......+ k^3 + (k + 1)^3 = k^2 (k + 1)^2/4 + (k + 1)^3
= (k + 1)^2[k^2/4 + (k + 1)]
= (k + 1)^2 ( k^2 + 4k + 4)/4
= (k + 1)^2(k + 2)^2/4
= (k +1)^2[(k + 1) + 1]^2/4 = R. H. S.