Let $0\xrightarrow{}E_0\xrightarrow{u_0}E_1\xrightarrow{u_1}E_2\xrightarrow{}\ldots\xrightarrow{}E_{n-1}\xrightarrow{u_{n-1}}E_n\xrightarrow{u_n}0$ be an exact sequence. Then $$\sum_{2k+1\leq n}\text{dim}(E_{2k+1})=\sum_{2k\leq n}\text{dim}(E_{2k}).$$
Write $F_k:=\text{Im}(u_k)$. Then $\text{dim}(E_{k+1})=\text{dim}(F_k)+\text{dim}(F_{k+1})$ for all $0\leq k\leq n-2$. I would like to show that both sides of the above equality are identical with $\sum_{i=0}^{n-1}\text{dim}(F_k)$.
The indexing of the sums is proving to be a challenge. How can I rewrite $\sum_{2k+1\leq n}\text{dim}(E_{2k+1})$ in terms of the $F_k$? Ideally I would like to reindex the first sum and then use that to show that the identity holds.