In the sequence $1,3,7,15,31\ldots$ each term is $2\cdot\text{immediately preceding term}+1$. What is the $n$-th term? I readily see that it is $2^n-1$, but how can I deduce the $n$-th term from the given pattern i.e. $n$-th term $= 2\cdot(n-1)\text{th term} + 1$ without computation.

Here is an explicit proof by induction. Given that $a_{n+1}=2a_n+1$ for $n=0,1,...$, with $a_0=1$ prove the proposition $\mathcal P(n):\;\,a_n=2^n-1$ for $n=0,1,...$ .
We are given that $\mathcal P(n)$ holds for $n=0$. Suppose now that we know $\mathcal P(n)$ for $n=k$; that is, $\mathcal P(k):\;\,a_k=2^k-1$. Now $a_{k+1}=2a_k+1=2(2^k-1)+1=2^{k+1}-1$. We have shown that $\mathcal P(k)\Rightarrow\mathcal P(k+1)$. Hence, by induction, $\mathcal P(n)$ holds for all $n=0,1,...$ .  
