Question

PROBLEM 6

Let Tn be the nth triangular number, Qn be the nth square number, and Pn be the nth pentagonal number.
(a) Show that Q6 + P5 = 3T5 +Q5 +1.
(b) Show that Qn+1 + Pn = 3Tn +Qn + 1 for all positive integers n.​

Answer 1

See below.

Explanation:

The nth triangular number is Tn = n(n + 1)/2, the nth square number Qn = n^2 and the nth pentagonal number Pn = (3n^2 - n) / 2.

(a. Q6 + P5 = 6^2 + (3(5)^2 - 5)/2

= 36 + 35 = 71.

3T5 + Q5 + 1 = 3* 5(5 + 1) / 2 + 5^2 + 1

= 90 / 2 + 26

= 71.

So Q6 + P5 = 3T5 +Q5 +1.

(b) Qn+1 + Pn = (n + 1)^2 + (3n^2 - n) / 2

= n^2 + 2n + 1 + (3n^2 - n) / 2

= n^2 + 2n + 1 + 1.5n^2 - 0.5n

= 2.5n^2 + 1.5n + 1.

3Tn +Qn + 1 = 3n(n + 1) / 2 + n^2 + 1

= 3n^2 + 3n / 2 + n^2 + 1

= 1.5n^2 + 1.5n + n^2 + 1

= 2.5n^2 + 1.5n + 1.

Therefore they are equal for all positive n.