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.

I know this is an old post, but I ended up having to solve the same question the other day, and initially looked here first.
Firstly we see the recursive formula $x_{n+1} = 2x_{n} + 1$. Letting our first term be $x_1$, then 
$$x_2 = 2x_1 + 1,$$
$$x_3 = 2x_2 + 1,$$
$$x_4 = 2x_3 + 1.$$
Rewriting these in terms of $x_1$ and looking at the powers of two yields:
$$x_2 = 2x_1 + 1 = 2^1x_1 + 1$$
$$x_3 = 4x_1 + 3 = 2^2x_1 + 3$$
$$x_4 = 8x_1 + 7 = 2^3x_1 + 7$$
It can be seen that $x_n = 2^{n-1}x_1 + x_{n-1}$. We already know that  $x_{n} = 2x_{n-1} + 1$. This can rearranged to show $x_{n-1} = \frac{x_{n}-1}{2}$
therefore:
$$x_n = 2^{n-1}x_1 + x_{n-1}$$
$$x_n = 2^{n-1}x_1 + \frac{x_n-1}{2}$$
$$2x_n = 2*2^{n-1}x_1 + x_n-1$$
$$x_n = 2^nx_1-1$$ And in our case $x_1 = 1$ therefore $$x_n = 2^n-1$$
