Difference Equation, verify expression is solution to the equation I am reading a book on Probability, and do not know how to solve this example question. Consider the following difference equation and initial condition(s). In each case, verify that the expression given for a(subscript(n)) (i) is a solution of the equation (ii) satisfies the initial condition(s). a(subscript(n)) = 16(1/2)^n satisfies 2a(subscript(n)) = a(subscript(n−1)), a(subscript(0)) = 16.

All you need to do here is substitute in $$16(1/2)^n ~~~~ \textrm{and} ~~~~ 16(1/2)^{n-1}$$ for $a_n$ and $a_{n-1}$, respectively, in the difference equation. You should see that the given choice of $a_n$ follows the rule provided (this is part (i)). Then check that, by plugging in $n = 0$, you indeed find $a_0 = 16$ (this is part (ii)).

Some students find the idea of verifying a solution like this is unintuitive; you start with both the question and the answer - what are you expected to do? It might help to see that another choice of function does $not$ satisfy the equation. Suppose you were told to do the same thing with $$ a_n = 16 + n$$ or $$ a_n = 7(1/2)^n.$$ These are examples of functions that don't solve the equation and initial condition.