Show that f(x) = x^n satisfies f(1)+ (f'(1))/(1!)+
Show that $f(x) = x^{n}$ satisfies \[ f(1)+\frac{f'(1)}{1!}+\frac{f''(1)}{2!}+\cdots+\frac{f^{(n)}(1)}{n!} = 2^{n}. \].
Explanation of Method
The Taylor coefficients at x = 1 are the binomial coefficients.
Step 1: The derivatives of $x^{n}$
From Problem 287, for $0\le k\le n$: $f^{(k)}(x) = \frac{n!}{\bigl(n-k\bigr)!}\,x^{\,n-k}$.
Step 2: Evaluate at $x = 1$
Every power of $1$ is $1$, so the whole $x$-dependence disappears: $f^{(k)}(1) = \frac{n!}{\bigl(n-k\bigr)!}$.
Step 3: Divide by $k!$
The quotient is exactly a binomial coefficient: $\frac{f^{(k)}(1)}{k!} = \frac{n!}{k!\,\bigl(n-k\bigr)!} = \binom{n}{k}$.
Step 4: Sum
\[ \sum_{k=0}^{n}\frac{f^{(k)}(1)}{k!} = \sum_{k=0}^{n}\binom{n}{k} = \bigl(1+1\bigr)^{n} = 2^{n}, \]
by the binomial theorem with both arguments equal to $1$.
Step 5: Verification
\[n = 1:\ 1+1 = 2 = 2^{1}; \qquad n = 2:\ 1+2+1 = 4 = 2^{2} \;\;\checkmark\]
\[n = 3:\ 1+3+3+1 = 8 = 2^{3}; \qquad n = 4:\ 1+4+6+4+1 = 16 \;\;\checkmark\]
\[\text{Each row of Pascal's triangle sums to } 2^{n} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\sum_{k=0}^{n}\frac{f^{(k)}(1)}{k!} = 2^{n}\,}\]
Remark
Remark. The sum is nothing but the Taylor expansion of $f$ about $x = 1$ evaluated at $x = 2$: since $f$ is a polynomial the expansion is exact and finite, and it gives $f(2) = 2^{n}$ directly. Read that way the identity is not a coincidence but a tautology — Taylor's formula for $x^{n}$ about $1$ is the binomial theorem for $\bigl(1+1\bigr)^{n}$.