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# Can anyone use mathematical induction to prove 1^2 + 4^2 + 7^2 +...+ (3n-2)^2 = 1/2n(6n^2-3n-1) without factorizing a cubic?

### 3 Answers

- PopeLv 71 month agoFavorite Answer
I suppose mathematical induction would be the way to go. Let P(n) represent this proposition:

1² + 4² + 7² + ... + (3n - 2)² = 1/2n(6n² - 3n - 1)

P(1):

1² = 1/2(1)[6(1)² - 3(1) - 1]

LHS = 1

RHS = 1

Proposition P(1) is confirmed.

Suppose P(k) is true for some positive integer k.

1² + 4² + 7² + ... + (3k - 2)² = 1/2k(6k² - 3k - 1)

P(k + 1):

1² + 4² + 7² + ... + (3k - 2)² + [3(k + 1) - 2]² = 1/2(k + 1)[6(k + 1)² - 3(k + 1) - 1]

LHS

= 1² + 4² + 7² + ... + (3k - 2)² + [3(k + 1) - 2]²

= 1/2k(6k² - 3k - 1) + [3(k + 1) - 2]²

= 1/2k(6k² - 3k - 1) + (3k + 1)²

= 1/2(6k³ - 3k² - k) + 9k² + 6k + 1

= 1/2(6k³ - 3k² - k) + 1/2(18k² + 12k + 2)

= 1/2(6k³ + 15k² + 11k + 2)

= 1/2(k + 1)(6k² + 9k + 2)

= 1/2(k + 1)(6k² + 12k + 6 - 3k - 3 - 1)

= 1/2(k + 1)[6(k + 1)² - 3(k + 1) - 1]

= RHS

So P(k + 1) is true.

It has been proved that P(1) is true, and if P(k) is true for any positive integer k, then P(k + 1) must also be true. Therefore, by mathematical induction, P(n) is true for all positive integers n.

- 1 month ago
You wanted this proved by induction without factoring a cubic. @Pope proved – correctly – that

(½ )(6k³ + 15k² + 11k + 2) =

(½ )((k + 1)[6(k + 1)² - 3(k + 1) – 1]

But he factored a cubic in his proof. The equality of these two expressions can be proved without such factoring as follows. If two polynomials of the same degree (say n) agree for n values of the variables, then they must be identical. Set k =1. Each side equals 17. For k = 0, they are each equal to 1 and for k = -1, they are each equal to 0. So the polynomials are identical. That is all you need.

(I can prove your result without using induction – by a technique known as the “method of differences” – but I assume induction was essential to your question).