Here is the question:
There are $m$ different sizes of disks and exactly $n_k$ disks of size $k$. Determine $A(n_l,. . . , n_m)$, the minimum number of moves needed to transfer a tower when equal-size disks are considered to be indistinguishable.
First, I figured out the number of moves needed for each disk when there are $n$ disks. I found out that each disk moves the same amount regardless the total number of disks, so $f(n)=t(n)-t(n-1)$.
$f(n)$ is the minimum number of times the disk moves, starting from the bottom and going up. For example: $f(1)$ is for the most bottom disk and $f(2)$ is the one above it etc.
$t(n)$ is the minimum number of moves needed to move all n disks from peg A to peg B.
After this, I just did sum from $k=1$ to $n$, the number of disks, of $(t(k)-t(k-1)) \cdot n_k$.
Is this the right answer?
It isn't the right answer because it isn't an answer. It is a plan to find the answer. Why do you talk of $A(n_l,. . . , n_m)$ when there are only three parameters in the problem-$m,k$ and $n$? I presume $k$ is the position in the list of sizes where you have multiple disks, so there are $m+n-1$ disks in total. Stating the question carefully is often a good step toward the solution.
Assume we are doing Tower of Hanoi with $m$ sizes of disk and there are $n$ disks of the $k^{\text{th}}$ smallest size. These disks can be stacked on on another in any order, but cannot be stacked on any smaller disk. Hint: you should convince yourself that all the matching disks need to stay together. Each time in the standard Tower of Hanoi that you would move a disk of size $k$, you make $n$ moves instead of $1$. How many times does that disk move in the standard Tower of Hanoi? The largest disk only moves once, the next moves ???