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$$0\to M\to Q_1\to Q_2\to\dots\to Q_i \to N\to 0$$ exact sequence, then $$H^n(N)\cong H^{n+i}(M)$$

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Standard trick: break up a long exact sequence into lots of little short exact sequences. –  Zhen Lin Nov 8 '12 at 20:15
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As stated this is simply false. To get an example, just take $i$ large and let $Q_2=\cdots=Q_{i-1}=0$. At the very least, you should tell us what $M$ and $N$ and so on are, and what cohomology you have in mind! –  Mariano Suárez-Alvarez Nov 8 '12 at 21:42
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You seem to have edited the question into something rather different to what it was at the time people wrote their answers. Please do not do that. –  Mariano Suárez-Alvarez Nov 8 '12 at 21:45

2 Answers 2

Use induction: Let $D_k$ be the image of $B_k \to B_{k+1}$. You have short exact sequences $A \rightarrowtail B_1 \twoheadrightarrow D_1$, $D_1 \rightarrowtail B_2 \twoheadrightarrow D_3$ up to $D_{m-1} \rightarrowtail B_m \twoheadrightarrow C$. From what you proved you know that $$ H^n(G,C) \cong H^{n+1}(G,D_{m-1}) \cong \cdots \cong H^{n+m-1}(G,D_1) \cong H^{n+m}(G,A) $$ for all $n \geq 1$ and you're done.

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Use the spectral sequence for the resulting bicomplex. If you use first the differential in the exact sequence, you see that the total cohomology vanishes. If you use first the differential for the group cohomology, then the 1st page of the spectral sequence contains $H^k(G,A)$'s in one column, and $H^k(G,C)$'s in another. As the total cohomology is $0$, the $m$'th differential (the one going between these columns) is an isomorphism, the one you look for.

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