Prove the identity sum_k=1ⁿ k4 = (n(n+1)(2n+1)(3n2+3n-1))/(30) using mathematical induction.

To prove the identity $\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ by mathematical induction, we follow these steps:

Step 1: Base Case ($n=1$)
Let $P(n)$ be the statement $\sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
For $n=1$:
LHS: $\sum_{k=1}^{1} k^4 = 1^4 = 1$.
RHS: $\frac{1(1+1)(2 \cdot 1+1)(3 \cdot 1^2+3 \cdot 1-1)}{30} = \frac{1(2)(3)(3+3-1)}{30}