I'm having a tough time proving the following formula. Suppose $a,b\in R$ a ring. Define $a^{(0)}=a$, $a^{(1)}=[a,b]\equiv ab-ba$, and then $a^{(k)}=[a^{(k-1)},b]$. Then $ \sum_{i=0}^k b^iab^{k-i}=\sum_{j=0}^k\binom{k+1}{j+1}b^{k-j}a^{(j)}. $
I wanted to do this with induction on $k$, but the presence of $k$ in each term of the sum is making it difficult to get the right form of things. I had something like $ \begin{align*} \sum_{i=0}^{k+1}\binom{k+2}{j+1}b^{k+1-j}a^{(j)} &= \sum_{j=0}^k \binom{k+2}{j+1}b^{k+1-j}a^{(j)}+a^{(k+1)}\\ &= \sum_{j=0}^k\left[\binom{k+1}{j}+\binom{k+1}{j+1}\right]b^{k+1-j}a^{(j)}+a^{(k+1)}\\ &=b\sum_{j=0}^k\binom{k+1}{j}b^{k-j}a^{(j)}+b\sum_{j=0}^k\binom{k+1}{j+1}b^{k-j}a^{(j)}+a^{(k+1)}\\ &=b\sum_{j=0}^k\binom{k+1}{j}b^{k-j}a^{(j)}+b\sum_{i=0}^k b^iab^{k-i}+a^{(k+1)} \end{align*} $ but this is still far from what I want. I tried doing induction in the other direction, but had similar problems. I don't see how the recursive definition of $a^{(k)}$ comes into play. How can the formula be properly derived? Thanks.