`1 + 4 + 7 + 10 + ... (3n - 2) = n/2 (3n - 1)` Use mathematical induction to prove the formula for every positive integer n.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:

Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:

`1 = 1/2*(3*1-1) => 1 =2/2 => 1=1`

Step 2:...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:

Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:

`1 = 1/2*(3*1-1) => 1 =2/2 => 1=1`

Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:

`P(k): 1 + 4 + 7 + .. + (3k-2) = (k(3k-1))/2 ` holds

`P(k+1):  1 + 4 + 7 + .. + (3k-2) + (3k+1) =  ((k+1)(3k+2))/2`

You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:

`(k(3k-1))/2 + (3k+1) = ((k+1)(3k+2))/2`

`3k^2 - k + 6k + 2 = 3k^2 + 2k + 3k + 2`

You need to add the like terms, such that:

`3k^2 + 5k + 2 = 3k^2 + 5k + 2`

Notice that P(k+1) holds.

Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement `P(n): 1 + 4 + 7 + .. + (3n-2) = (n(3n-1))/2`  holds for all positive integers n.

Approved by eNotes Editorial Team