Final answer:
The equation 3^n = sum of [n; k] * 2^k for k=0 to n is proven using the binomial theorem, which describes the expansion of (a + b)^n when a=1 and b=2, thus confirming the given expression for any positive integer n.
Step-by-step explanation:
We are asked to prove that for any positive integer n, the equation 3n = ∑k=0n([ n; k ]) 2k holds true. This expression represents the expansion of a binomial raised to an exponent, specified by the binomial theorem.
To prove this equality, let's consider the binomial theorem which states that the expansion of (a + b)n is given by an + n an-1b + n(n-1)/2! an-2b2 + ... + bn. If we choose a = 1 and b = 2, we get (1 + 2)n = 3n, which coincides with the left side of our original equation.
By expanding (1 + 2)n according to the binomial theorem, we can see that each term of the form n! / (k! (n-k)!) * 1n-k * 2k will be present. The term n! / (k! (n-k)!) is the binomial coefficient, denoted as [ n; k ]. Therefore, each term in the expansion represents a term in the summation ∑k=0n([ n; k ]) 2k, confirming that our original equation is correct.