SOLUTION: use mathematical induction to prove that the statement is true for all positive integers. 5+23+53+...6n^2 - 1=n^2 (2n+3)

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Question 1115661: use mathematical induction to prove that the statement is true for all positive integers.
5+23+53+...6n^2 - 1=n^2 (2n+3)
Found 2 solutions by t0hierry, greenestamps:
Answer by t0hierry(194) About Me  (Show Source):
You can put this solution on YOUR website!
The Sum S is equal to 6 Sum i^2 - n
6 Sum i^2 -n
= n(n+1)(2n+1) - n
= n(2n^2 + 3n + 1) - n
= n(2n^2 + 3n)
= n^2(2n + 3)

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


To prove by mathematical induction:
(1) show the statement is true for some beginning value (usually 1, but not always); and
(2) show that, assuming it is true for some integer n, it follows that it is true for n+1 also

The statement is true for n=1: 5 = (1^2)(2(1)+3) = 1*5 = 5

We need to show that if

5+23+53+...+(6n^2-1) = n^2(2n+3)

is true, then it follows algebraically that

5+23+53+...+(6n^2-1)+(6(n+1)^2-1) = (n+1)^2(2(n+1)+3)

On the left, replace the sum up to the (6n^2-1) term with the expression n^2(2n+3) and simplify. And simplify the expression on the right.

The proof is complete if the two expressions are the same.

On the left...
n%5E2%282n%2B3%29%2B%286%28n%2B1%29%5E2-1%29 =
2n%5E3%2B3n%5E2%2B6%28n%5E2%2B2n%2B1%29-1 =
2n%5E3%2B3n%5E2%2B6n%5E2%2B12n%2B6-1 =
2n%5E3%2B9n%5E2%2B12n%2B5

On the right...
%28n%2B1%29%5E2%282%28n%2B1%29%2B3%29 =
%28n%5E2%2B2n%2B1%29%282n%2B5%29 =
2n%5E3%2B5n%5E2%2B4n%5E2%2B10n%2B2n%2B5 =
2n%5E3%2B9n%5E2%2B12n%2B5

The proof by mathematical induction is complete.