1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = quantity n times quantity six n squared minus three n minus one all divided by two

I assume you're trying to prove this identity for natural numbers n?

Let P(n) be the statement that

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

We prove it by induction. First check that P(1) is true:

1² = 1

1 • (6•1² - 3•1 - 1) / 2 = 1 • 2 / 2 = 1

so P(1) is indeed true.

Assume P(k) is true, that

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

We want to use this hypothesis to show that P(k + 1) is also true, that

1² + 4² + 7² + … + (3k - 2)² + (3 (k + 1) - 2)²

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

= (k + 1) (6k² + 9k + 2) / 2

= (6k³ + 15k² + 11k + 2) / 2

By the induction hypothesis,

1² + 4² + 7² + … + (3k - 2)² + (3 (k + 1) - 2)²

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

= k (6k² - 3k - 1) / 2 + (3k + 1)²

= 3k³ + 15/2 k² + 11/2 k + 1

= (6k³ + 15 k² + 11 k + 2) / 2

which is exactly what we needed to show, so P(k + 1) is also true.

Thus P(n) is true for all natural numbers n.

hope this .

r3t40

16.9 times 2.85

step-by-step explanation: