The given series is:
3(2x+1) + 3(2x+1) + 3(2x+1) + ... + 3(2x+1) (adding up to n terms)
We can simplify this by factoring out the common factor of 3(2x+1) from each term, which gives:
3(2x+1)(1 + 1 + 1 + ... + 1) (adding up to n terms)
The expression in parentheses represents the sum of n ones, which is simply n. Therefore, we have:
3(2x+1)(n)
This is the formula for the sum of the given series.
To verify this formula, we can use mathematical induction.
Base case: When n=1, the sum is 3(2x+1)(1) = 6x+3, which is the first term of the series. So the formula holds for n=1.
Inductive step: Assume that the formula holds for some integer k, i.e., the sum of the first k terms is 3(2x+1)(k).
We want to show that the formula also holds for k+1, i.e., the sum of the first k+1 terms is 3(2x+1)(k+1).
Adding the (k+1)th term 3(2x+1) to the sum of the first k terms, we get:
3(2x+1)(k) + 3(2x+1)
Factoring out the common factor of 3(2x+1), we have:
3(2x+1)(k+1)
This is the formula for the sum of the first k+1 terms.
Therefore, by mathematical induction, we have verified that the formula 3(2x+1)(n) is correct for the sum of the given series.