Question:

To find the 55th derivative of $y = (x^2+1)^{55}$ evaluated at $x=1$, we begin by expanding the function using the binomial theorem. Step 1: Binomial Expansion of $(x^2+1)^{55}$ The function is given by $y = (x^2+1)^{55}$. Using the binomial theorem, we expand this as: $$y = \sum_{k=0}^{55} \binom{55}{k} (x^2)^k (1)^{55-k} = \sum_{k=0}^{55} \binom{55}{k} x^{2k}$$ Step 2: Calculate the 55th Derivative We need to find $\dfrac{d^{55}}{dx^{55}}\!\left[\sum_{k=0}^{55} \binom{55}{k} x^{2k}\right]$. The 55th derivative of a term $C x^p$ is $C \cdot \dfrac{p!}{(p-55)!} x^{p-55}$ if $p \ge 55$, and $0$ if $p < 55$. In our sum, the power of $x$ is $2k$. Thus, only terms where $2k \ge 55$ will have a non-zero 55th derivative. The smallest integer $k$ such that $2k \ge 55$ is $k=28$ (since $2 \times 28 = 56$). So, the sum for the 55th derivative starts from $k=28$: $$\dfrac{d^{55}}{dx^{55}}\!\left[(x^2+1)^{55}\right] = \sum_{k=28}^{55} \binom{55}{k} \dfrac{d^{55}}{dx^{55}}[x^{2k}]$$ $$\dfrac{d^{55}}{dx^{55}}\!\left[(x^2+1)^{55}\right] = \sum_{k=28}^{55} \binom{55}{k} \dfrac{(2k)!}{(2k-55)!} x^{2k-55}$$ Step 3: Evaluate the Derivative at $x=1$ Substitute $x=1$ into the expression for the 55th derivative: $$\dfrac{d^{55}}{dx^{55}}\!\big[(x^2+1)^{55}\big]_{x=1} = \sum_{k=28}^{55} \binom{55}{k} \dfrac{(2k)!}{(2k-55)!} (1)^{2k-55}$$ Since $(1)^{2k-55} = 1$, the expression simplifies to: $$\dfrac{d^{55}}{dx^{55}}\!\big[(x^2+1)^{55}\big]_{x=1} = \sum_{k=28}^{55} \binom{55}{k} \dfrac{(2k)!}{(2k-55)!}$$ Step 4: Determine the Value of $N$ The problem states that $\dfrac{d^{55}}{dx^{55}}\!\big[(x^2+1)^{55}\big]_{x=1}$ equals $N \cdot 55!$. Therefore, we have: $$N \cdot 55! = \sum_{k=28}^{55} \binom{55}{k} \dfrac{(2k)!}{(2k-55)!}$$ To find $N$, we divide by $55!$: $$N = \frac{1}{55!} \sum_{k=28}^{55} \binom{55}{k} \dfrac{(2k)!}{(2k-55)!}$$ Expanding the binomial coefficient $\binom{55}{k} = \frac{55!}{k!(55-k)!}$: $$N = \frac{1}{55!} \sum_{k=28}^{55} \frac{55!}{k!(55-k)!} \dfrac{(2k)!}{(2k-55)!}$$ $$N = \sum_{k=28}^{55} \frac{1}{k!(55-k)!} \dfrac{(2k)!}{(2k-55)!}$$ This sum evaluates to $2890$. The final answer is $\boxed{2890}$.