Question 1184006
I will go directly to step 3 of the Inductive process, and will assume the Inductive hypothesis to be true, i.e., 

{{{1^4 + 2^4 + 3^4 + "..." + k^4 = (1/30)k(k+1)(2k+1)(3k^2 + 3k - 1)}}}  for some n = k.

Prove: {{{1^4 + 2^4 + 3^4 + "..." + k^4 = (1/30)k(k+1)(2k+1)(3k^2 + 3k - 1)}}} is true for n = k + 1, that is, 

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


From the inductive hypothesis, 
{{{1^4 + 2^4 + 3^4 + "..." + k^4 + (k+1)^4  = (1/30)k(k+1)(2k+1)(3k^2 + 3k - 1) + (k+1)^4 = ((k+1)/30)*red((k(2k+1)(3k^2 + 3k - 1) + 30(k+1)^3))}}}


Focus on the the polynomial {{{red(p(k)) = k(2k+1)(3k^2 + 3k - 1) + 30(k+1)^3}}} , which is of 4th degree.

By the factor theorem, {{{k+2}}} is a factor of this polynomial since {{{p(-2) = -2*(2(-2)+1)(3(-2)^2 + 3(-2) - 1) + 30((-2)+1)^3 = 0}}}.

Again by the factor theorem, {{{2k+3}}} is a factor of this polynomial since {{{p(-3/2) = (-3/2)*(2(-3/2)+1)(3(-3/2)^2 + 3(-3/2) - 1) + 30((-3/2)+1)^3 = 0}}}.

In other words, {{{red(p(k)) = red( (k+2)(2k+3)p[2](k))}}} , where {{{p[2](k) = ak^2+bk +c}}} is a quadratic expression.

Now {{{red(p(0)) = 0(2*0+1)(3*0^2 + 3*0 - 1) + 30(0+1)^3 =red( (0+2)(2*0+3)(a*0^2 + b*0 +c))}}}   ===>   {{{30 = 6c}}}  ===> {{{c=5}}}.

Also, {{{p(k)}}} is a 4th degree polynimial whose leading term is {{{6k^4}}}.  Since  {{{p(x) = (k+2)(2k+3)(ak^2+bk +c)}}} also, this means {{{a = 3}}}.

===> {{{p(k) = k(2k+1)(3k^2 + 3k - 1) + 30(k+1)^3 = red( (k+2)(2k+3)(3k^2+bk+5))}}} 


===> {{{p(1) = 1*3(3*1^2 + 3*1 - 1) + 30(1+1)^3 = red( (1+2)(2*1+3)(3*1^2+b*1+5))}}} ===> {{{3*5 +30*8 = 3*5*(b+8)}}}  ===>  {{{b=9}}}.

===> {{{p(k) = red( (k+2)(2k+3)(3k^2+9k+5)) = (k+2)(2k+3)(3(k+1)^2+3(k+1)-1)}}}.


Therefore,  {{{1^4 + 2^4 + 3^4 + "..." + k^4 + (k+1)^4  = (1/30)k(k+1)(2k+1)(3k^2 + 3k - 1) + (k+1)^4 = ((k+1)/30)*red((k(2k+1)(3k^2 + 3k - 1) + 30(k+1)^3)) = ((k+1)/30)* (k+2)(2k+3)(3(k+1)^2+3(k+1)-1)  }}}, 

and the proof is complete.