Question 1184006
prove using mathematical induction:

1. 1^4 + 2^4 + 3^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1)

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Base case, n=1:   LHS = 1^4 = 1
                  RHS = (1/30)(1*2*3*5) = (1/30)(30) = 1   (base case holds) 


Hypothesis:
Assume {{{ 1^4 + 2^4 + 3^4 }}} + ... + {{{n^4}}} = {{{ (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1) }}}   for  n=k


Step case:  Let n=k+1
Thus far, it has been setup.  The task now is to show LHS = RHS for n=k+1, and the proof will be complete.

LHS = {{{ 1^4 + 2^4 + 3^4 }}} + ... + {{{ (k+1)^4 }}} 

...which can also be written...

= {{{ green( 1^4 + 2^4 + 3^4) }}} + ... + {{{ green(k^4) }}} + {{{ (k+1)^4 }}} 

...apply the hypothesis to {{{green(green)}}} terms...

= {{{ (1/30)k(k+1)(2k+1)(3k^2 + 3k - 1) }}} + {{{ (k+1)^4 }}} 


...expand and simplify (lots of steps omitted here)...
= {{{ (1/30)(k+1)(k+2)(2k+3)(3k^2+9k+5) }}}

Proof is complete, but lets show this last expression is of the form of the RHS {{{ (1/30)n(n+1)(2n+1)(3n^2 + 3n - 1) }}} ...

Let u = k+1  -->  k=u-1:

LHS =  {{{ (1/30)(u)(u+1)(2(u-1)+3)(3(u-1)^2+9(u-1)+5) }}}
    =  {{{ (1/30)u(u+1)(2u+1)(3u^2+3u-1) }}}  (= RHS)