2^n(n+1) for n>=0 mathematical induction
To prove the statement 2^n(n+1) for n≥0 by mathematical induction, we first need to establish a base case and then show the inductive step.
1. Base case:
When n = 0, we have 2^0(0+1) = 2^0 = 1.
So the statement is true for n = 0.
2. Inductive step:
Assume that the statement is true for some positive integer k. That is, assume 2^k(k+1) holds.
Now we need to prove that the statement holds for (k+1).
Using the assumption, we have 2^k(k+1) = 2^k * k + 2^k.
Now let's evaluate 2^(k+1)((k+1)+1):
2^(k+1)((k+1)+1) = 2^(k+1)(k+2)
Expanding the expression:
= (2^k * 2)((k+1) + 1)
= 2^(k+1)(k+2)
= 2^k * 2 * (k+2)
= 2^k(k+1) * 2 + 2^k * 2
= 2^k(k+1) + 2^k * 2 (By the assumption)
Now, 2^k(k+1) + 2^k * 2 is equal to the assumption, which is 2^k(k+1), plus an additional term 2^k * 2.
Since the assumption is true, we have 2^k(k+1) + 2^k * 2 = 2^k(k+1) + 2^(k+1).
Therefore, we have shown that if the statement is true for some positive integer k, then it is also true for (k+1).
By the principle of mathematical induction, the statement 2^n(n+1) holds for n≥0.