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# Show that p_k = p_(k-1) + (3k-2) for k >=2. Conclude that p_n = sum from k=1 to n of (3k-2).?

where p_k is the kth pentagonal number

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- 1 decade agoFavorite Answer
Let k ≥ 2. Then

p_(k - 1) + 3k - 2 =

(k - 1)[3(k - 1) - 1]/2 + (6k - 4)/2 =

[(k - 1)(3k - 3 - 1) + 6k - 4]/2 =

[(k - 1)(3k - 4) + 6k - 4]/2 =

(3k² - 7k + 4 + 6k - 4)/2 =

(3k² - k)/2 =

k(3k - 1)/2 = p_k.

Now consider the sum

Σ (k = 2 to n) (p_k - p_(k - 1)).

It is a telescoping sum equal to p_n - p_1 = p_n - 1. As shown above, we see that each summand p_k - p_(k - 1) = 3k - 2. It follows that

p_n = Σ (k = 2 to n) (3k - 2) + 1 =

Σ (k = 2 to n) (3k - 2) + 3(1) - 2 =

Σ (k = 1 to n) (3k - 2).

I hope that helps!

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