Question

Use mathematical induction to prove the statement is true for all positive integers n.

Answer 1

First verify the base case by inserting values so you can assume it holds for it n

for n+1 add the next term 6*(n+1) of the sum to both sides:

6 + 12 + 18 + ... + 6n +6(n+1)= 3n(n+1)+6(n+1)
=(n+1)(3n+6)
=3(n+1)(n+2)
=3(n+1)((n+1)+1)
=term if you had inserted n+1 instead of n, proving correctness

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Answer 2

Answer:

Step-by-step explanation:

The given statement is:

6 + 12 + 18 + ... + 6n = 3n(n + 1)

we have t prove that the given statement is true for all positive integers n.

Thus, using mathematical induction, we have

Step 1: Let us show that the statement is true for n = 1.  

Putting n=1 in the given statement, we have

6 = 3(1)(1 + 1). And this is clearly a true statement.

So the statement is true for n = 1.  

Step 2: Show if it holds for k, then it holds for k + 1.  

Let us assume that the statement is true for a positive integer k. That is,

6 + 12 + ... + 6k = 3k(k + 1).  

Let us show that the statement must be true for the positive integer k + 1. We need to show that the sum of the first k + 1 multiples of 6 is equal to 3(k+1)( (k + 1) + 1).  

Thus,

6 + 12 + ... + 6k + 6(k+1) = [6 + 12 + ... + 6k] + 6(k + 1)  

= 3k(k + 1) + 6(k + 1) (this is true by the induction hypothesis)  

= 3(k + 1) [ k + 2 ] (Factorizing 3(k + 1) )  

= 3(k + 1)[(k + 1) + 1]

Therefore, the result holds.

Hence, the given statement is true for all positive integers n.