Let
$$0 \overset{d_{n+1}}{\rightarrow} G_n \overset{d_n}{\rightarrow} G_{n-1}\overset{d_{n-1}}{\rightarrow}\ldots\overset{d_2}{\rightarrow}G_1 \overset{d_1}{\rightarrow} 0$$ be a sequence of modules and homomorphisms over the commuative ring $R$ such that $d_i \circ d_{i+1}=0$ for all $i=1,..,n-1$, and suppose that $G_i$ has finite length for all $i$.
Set $H_i=\ker d_i / \text{im} \, d_{i+1}$. Show that $H_i$ has finite length for all $i$ and that $$\sum_{i=0}^{n} (-1)^i l(H_i)=\sum_{i=0}^{n} (-1)^i l(G_i)$$
It's obvious to see that $H_i$ has finite length by the exact sequence $$0 \to Im \; d_{i+1}\to Ker \; d_i \to Ker \; d_i / Im \; d_{i+1}\to 0$$ But I can indicate that $\sum_{i=0}^{n} (-1)^i l(H_i)=\sum_{i=0}^{n} (-1)^i l(G_i)$. Can anyone help me?
I think $R$ can be an arbitrary ring (not necessarily unital). The fact that each $H_i$ is of finite length is due to the Jordan-Hölder theorem. I assume that $l$ is the module length function. We first prove a general result (1).
Let $A\overset{\alpha}{\longrightarrow}B\overset{\beta}{\longrightarrow}C$ be a sequence of homomorphisms of (not necessarily unitary) left $R$-module, where $A$, $B$, and $C$ are left $R$-modules of finite length and $\beta\circ\alpha=0$. Write $B'$ for $\ker\beta/\operatorname{im}\alpha$, and $q:\ker \beta\to B'$ is the canonical projection. We can decompose the maps into short exact sequences of left $R$-modules as follows: $$0\to\ker \alpha \overset{\subseteq}{\hookrightarrow} A\overset{\alpha}{\twoheadrightarrow} \operatorname{im}\alpha\to 0,$$ $$0\to \operatorname{im}\alpha \overset{\subseteq}{\hookrightarrow} \ker\beta \overset{q}{\twoheadrightarrow} B' \to 0,$$ and $$0\to \ker\beta\overset{\subseteq}{\hookrightarrow}B \overset{\beta}{\twoheadrightarrow} \operatorname{im}\beta\to 0.$$ It is easy to see that $$l(B)=l(\ker\beta)+l(\operatorname{im}\beta),$$ $$l(\ker\beta)=l(\operatorname{im}\alpha)+l(B'),$$ and $$l(\operatorname{im}\alpha)=l(A)-l(\ker\alpha).$$ Thus, $$l(B)=l(B')+l(A)-l(\ker\alpha)+l(\operatorname{im}\beta).\tag{1}$$
Now, if you have $0 \overset{d_{n+1}}{\rightarrow} G_n \overset{d_n}{\rightarrow} G_{n-1}\overset{d_{n-1}}{\rightarrow}\ldots \overset{d_2}{\rightarrow}G_1 \overset{d_1}{\rightarrow} 0$, then for $i=0,1,2,\ldots,n$, we have from (1) that $$l(G_i)=l(H_i)+l(G_{i+1})-l(\ker d_{i+1})+l(\operatorname{im}d_i).$$ Here, $G_0=G_{n+1}=0$, $d_0=0$, and $H_0=0$. Thus, $$l(G_i)-l(G_{i+1})=l(H_i)-l(\ker d_{i+1})+l(\operatorname{im}d_i).$$ That is, $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)-\sum_{i=0}^{n}\,(-1)^il(\ker d_{i+1})+\sum_{i=0}^n(-1)^il(\operatorname{im}d_i).$$ So, $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)+\sum_{i=1}^{n+1}\,(-1)^il(\ker d_{i})-\sum_{i=1}^{n+1}(-1)^{i}l(\operatorname{im}d_{i-1}).$$ Since $\operatorname{im}d_{0}=\operatorname{im}d_1=0$ and $\ker d_{n+1}=\operatorname{im}d_{n+1}=0$, we get $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)+\sum_{i=1}^{n}\,(-1)^il(\ker d_{i})-\sum_{i=3}^{n+2}(-1)^{i}l(\operatorname{im}d_{i-1}).$$ Thus, $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)+\sum_{i=1}^{n}\,(-1)^il(\ker d_{i})-\sum_{i=1}^{n}(-1)^{i}l(\operatorname{im}d_{i+1}).$$ That is, $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)+\sum_{i=1}^{n}\,(-1)^i\big(l(\ker d_{i})-l(\operatorname{im}d_{i+1})\big).$$ Because $l(H_i)=l(\ker d_{i})-l(\operatorname{im}d_{i+1})$, we obtain $$\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=\sum_{i=0}^n (-1)^i l(H_i)+\sum_{i=1}^{n}\,(-1)^il(H_i)=2\sum_{i=1}^{n}\,(-1)^il(H_i).$$ It is not difficult to see that $\displaystyle\sum_{i=0}^n(-1)^i\big(l(G_i)-l(G_{i+1})\big)=2\sum_{i=1}^{n}\,(-1)^il(G_i)$. This means $$\sum_{i=1}^{n}\,(-1)^il(G_i)=\sum_{i=1}^{n}\,(-1)^il(H_i).$$
As a warning, maybe we should remove the trivial cases $n=0$ (with the sequence $0\to 0$), $n=1$ (with the sequence $0\to G_1\to 0$), and $n=2$ (with the sequence $0\to G_2\to G_1\to 0$ from the proof above. Deal with them separately, but it is pretty easy for these cases.