= Solution
The finite form of <Taylor's theorem> with the first $n$ terms is
$$
\boxed{f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x),\qquad R_n(x)=\frac{f^{(n)}(\xi)}{n!}(x-a)^n,}
$$
where $\xi$ lies strictly between $a$ and $x$ for $x\ne a$. A sufficient hypothesis is that $f$ has continuous <derivatives> through order $n-1$ on the interval joining the endpoints and an $n$th <derivative> in its interior. The displayed polynomial has degree at most $n-1$; its <Taylor remainder> uses the $n$th <derivative>, not the $(n+1)$st. At $x=a$ the remainder is zero. This is a finite <Taylor polynomial> identity and does not require an infinite <Taylor series> to converge.
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