It looks like there were just some arithmetic errors. We see that
\begin{align}
431 &= 26 \cdot 16 &+ 15 \\
26 &= 1 \cdot 16 &+ 10 \\
1 &= 0 \cdot 16 &+ 1
\end{align}
Reading the remainders upward from the bottom, we have $(1\ 10 \ 15)_{16}$, commonly written as $(1AF)_{16}$.
As noted in the comments, your binary expansion has one too many ones; it should be $(110101111)_2$. For fun, you can get from binary to/from hex directly:
\begin{align}
(110101111)_2 &= (1\ 1010\ 1111)_2 \\
&= (1 \ A \ F)_{16}
\end{align}
so you could potentially use that as a sanity check.
This is a good start, but if I understand your method correctly, it does not quite work. It proves that you can represent infinitely many numbers— but it does not clearly show that you can represent all numbers.
For contrast, consider a similar claim:
You can represent any integer using a binary number whose unit bit is 0.
This claim is false because if the unit bit is 0, you can only represent even numbers. However, according to your proposed proof strategy, you might argue: (a) in the base case, you have one bit and can represent one number. (b) if you have $(k+1)$ bits, you can simply add $2^k$ to each of the numbers you have represented so far to get twice as many new numbers. This is true, but it only implies that you can represent infinitely many numbers in this scheme— in fact, you can represent all the even numbers. It does not imply that you can represent all integers, however. And indeed you can't represent odd numbers this way.
If you like, you could treat your goal formally as establishing a bijection between the natural numbers $\mathbb{N}$ and binary strings $\mathbb{B}$.
Thus, you would need to show that every integer has a bitstring representation, and that bitstring representations are unique— two different bitstrings must correspond to different numbers. (I'm assuming that leading zeroes are forbidden.)
But perhaps you only want to prove that every integer has a corresponding bitstring representation, and you don't care about uniqueness.
This is easy enough to show— we just convert each number to binary! Let $n$ be a non-negative integer.
Put $n_0 = n$ and $b_0 = n\pmod{2}$. Until $n_k$ is zero, define $n_{k+1} = (n_k - b_k)/2$ and define $b_{k+1} = n_{k+1}\pmod{2}$.
Because the $n_k$ form a decreasing sequence of non-negative numbers, this process must eventually end. When it does, declare the bitstring of $n$ to be $b_k\ldots b_0$. This is the binary representation of $n$. Because you can apply this procedure to any number $n$, it proves that every number has a binary representation.
This proof does not show a few other things, namely that bitstrings are unique (each number has at most one representation), or that every bitstring corresponds to a valid number.
Best Answer
Well, since there are only $63$ natural numbers less than $64$, one way you could proceed is by brute force. These are the numbers that can be represented in binary by six bits. There are therefore only $\binom{6}{3} = 20$ different numbers that qualify. Write them all out, and then add them.
A slightly more clever way to proceed is to consider how many numbers have a $1$ in the most significant binary (MSB) position—the one corresponding to decimal $32$. That leaves two bits to go in the remaining five positions. There are $\binom{5}{2} = 10$ of those, so the MSB position contributes $320$. That isn't the sum of those ten numbers, but it is the sum of the contribution of that bit from those ten numbers.
Similarly, there are ten qualifying numbers that contain the second most significant bit, so they contribute $160$, and so on to the least significant bit. What is the total contribution?