Final answer:
To prove the given equation, we can use a telescoping series and rewrite the sum in terms of a geometric series. The proof shows that the sum can be expressed as (n − 1)2^(n+1) + 2.
Step-by-step explanation:
To prove that for every positive integer n, ∑_(k=0)^n k*2^k = (n − 1)2^(n+1) + 2, we can use a telescoping series. Let's start by expanding the sum:
∑_(k=0)^n k*2^k = 0*2^0 + 1*2^1 + 2*2^2 + ... + (n-1)*2^(n-1) + n*2^n
We can rewrite this as:
(0*2^0 + 1*2^1 + 2*2^2 + ... + (n-1)*2^(n-1) + n*2^n) + [(n-1)*2^n + n*2^n]
Simplifying further, we get:
(0 + 2 + 8 + ... + 2^(n-2) + 2^(n-1)) + 2^n(n-1 + n)
The first part of the sum can be expressed as a geometric series:
0 + 2 + 8 + ... + 2^(n-2) + 2^(n-1) = 2^n - 2
Substituting this back into the equation, we get:
(2^n - 2) + 2^n(n-1 + n) = 2^n - 2 + 2^n(2n-1) = 2^n(2n - 1) - 2
Finally, we can rewrite this as:
2^(n+1)(n-1) + 2