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

to prove the given result by mathematical induction method, first we will show that it is true for n=1
then we will assume that it is true for n = k, and with the help of assumption we will try to prove that the given result is true for n = k+1.
step I:
P(1) = 1
RHS = $\frac{1*2*(2+1)*(3+3-1)}{30}=\frac{2*3*5}{30}=1$
thus the given equation holds for n =1
step II:
let us assume that given equation holds for n = k
therefore
$1^4+2^4+.......+k^4=\frac{k(k+1)(2k+1)(3k^2+3k-1)}{30} ............\left(1\right)$
step III:
now let us find the sum for n = (k+1)
$$\begin{gathered} 1^4+2^4+......+k^4+(k+1{)}^4 \\ =\frac{k(k+1)(2k+1)(3k^2+3k-1)}{30}+(k+1{)}^4 \\ =\frac{(k+1)}{30}.\left[k\right(2k+1\left)\right(3k^2+3k-1)+30(k+1{)}^3] \\ =\frac{(k+1)}{30}.\left[6k^4+6k^3-2k^2+3k^3+3k^2-k+30.\{k^3+3k^2+3k+1\}\right] \\ =\frac{(k+1)}{30}.\left[6k^4+9k^3+k^2-k+30k^3+90k^2+90k+30\right] \\ =\frac{(k+1)}{30}.\left[6k^4+39k^3+91k^2+89k+30\right] \end{gathered}$$
now we will find the factor of $g\left(k\right)=6k^4+39k^3+91k^2+89k+30$
$$\begin{gathered} g(-2)=6*(-2{)}^4+39*(-2{)}^3+91*(-2{)}^2+89*(-2)+30 \\ =96-312+364-178+30 \\ =490-490 \\ =0 \end{gathered}$$
thus $(k+2)$ is a factor of g(k).
$$\begin{gathered} 6k^4+39k^3+91k^2+89k+30 \\ = 6k^3 (k+2)+27k^2(k+2)+37k(k+2)+15(k+2) \\ =6k^4+12k^3+27k^3+54k^2+37k^2+74k+15k+30 \\ g\left(k\right)=(k+2)(6k^3+27k^2+37k+15) \end{gathered}$$
now k = -3/2 is a factor of g(k).
thus by the factorization we can write:
$g\left(k\right)=(k+2)(2k+3)(3k^2+9k+5)$
therefore
$$\begin{gathered} 1^4+2^4+3^4+...........+k^4+(k+1{)}^4 \\ =\frac{(k+1)}{30}.(k+2\left)\right(2k+3\left)\right(3k^2+9k+5) \\ =\frac{(k+1)}{30}.\{(k+1)+1\}.\{2.(k+1)+1\}.\{3.(k+1{)}^2+3(k+1)-1\} \end{gathered}$$
thus the given equation also holds for n = k+1.
thus it is true for all integer values of n.

hope this helps you