In theory, there are other ways to check the result without re-generating it, using modular arithmetic. In practice, these methods probably won't be useful at all, but I'll outline what I'm talking about anyway.
Remember the "cast out 9's" method for checking your arithmetic? From a mathematical perspective, "casting out 9's" amounts to reducing each number modulo 9. Then, we check whether $(A \bmod 9) + (B \bmod 9) = (C \bmod 9)$. If the result was generated correctly (i.e., if $A+B=C$), this check will always succeed. If the result was generated incorrectly (i.e., if $A+B \ne C$), then the check might fail.
We can generalize this, as follows. Pick a random $k$-bit prime, call it $p$. Reduce $A$, $B$, and $C$ modulo $p$. Now, check whether $(A \bmod p) + (B \bmod p) = (C \bmod p)$. If this check fails, then we're sure that $C$ was generated incorrectly. If this check succeeds, then with high probability $C$ was generated correctly. In particular, if we have $A,B,C$ such that $A+B \ne C$, then with high probability (over the choice of $p$) we'll have $(A \bmod p) + (B \bmod p) \ne (C \bmod p)$.
How high is the probability of detecting an error? Basically, it is exponentially small in $k$. The probability can be bounded explicitly using the fact that an error goes undetected iff $A+B-C$ is a multiple of $p$ and using an upper bound on the number of $k$-bit prime divisors that $A+B-C$ can have (at most $2 \log n / \log k$, where $n$ is the number of bits in $A,B,C$) and using the prime number theorem (there are at least $2^k/k$ $k$-bit prime numbers). You end up with a probability that's at most something like $(2 k \log n / \log k)/2^k$, and this is asymptotically exponentially decreasing in $k$.
In particular, if you want the probability of an error going undetected to be at most $1/2^t$, it suffices to choose $k=O(t \log t \log \log n)$. In practice, if $t \ge 64$, the chance of an undetected error is far smaller than the chance of a cosmic ray messing everything up. So, if we think of $t$ as constant, then we need a $k=O(\log \log n)$-bit prime.
Is this faster than just doing the addition? I doubt it.