Question

Problem 6

Let T, be the nth triangular number, Qn be the nth square number and
Pn be the nth pentagonal number.
(a) Show that 3Pn = T3n-1.
(b) Show that Pn - Qn = Tn-1 and hence that Pyn - 3Pn = Qan.

Answer 1

The given symbol for the nth triangular number = Tₙ

The given symbol for the nth square number = Qₙ

The given symbol for the nth pentagonal number = Pₙ

The mathematical formula for the nth triangular number, Tₙ = n·(n + 1)/2

The nth square number is given by Qₙ = n²

The nth pentagonal number is given by Pₙ = n·(3·n - 1)/2

(a) T₃.ₙ₋₁ = (3·n - 1)·(3·n - 1 + 1)/2 = (3·n - 1)·(3·n)/2 = 3·(3·n - 1)·(n)/2 = 3·Pₙ

Therefore;

T₍₃.ₙ ₋ ₁₎ = 3·Pₙ

(b) Pₙ - Qₙ = n·(3·n - 1)/2 - n²

Expanding the above equation gives;

n·(3·n - 1)/2 - n² = (3·n² - n - 2·n²)/2 = (n² - n)/2 = n·(n - 1)/2

T₍ₙ ₋ ₁₎ = ((n - 1)· ((n - 1) + 1))/2 = ((n - 1)· n)/2 = n·(n - 1)/2

∴ Pₙ - Qₙ = n·(3·n - 1)/2 - n² = n·(n - 1)/2 = T₍ₙ ₋ ₁₎

∴ Pₙ - Qₙ = T₍ₙ ₋ ₁₎

Therefore, we get;

P₃.ₙ - Q₃.ₙ = T₍₃.ₙ ₋ ₁₎

Where, T₍₃.ₙ ₋ ₁₎ = 3·Pₙ from (a) above, gives;

P₃.ₙ - Q₃.ₙ = 3·Pₙ

∴ P₃.ₙ - 3·Pₙ = Q₃.ₙ

Plugging in the values, gives;

P₃.ₙ = (3·n)·(3·(3·n) - 1)/2 = (3·n)·((9·n) - 1)/2

3·Pₙ = 3·n·(3·n - 1)/2

P₃.ₙ - 3·Pₙ = (3·n)·((9·n) - 1)/2 - 3·n·(3·n - 1)/2 = (3·n)·(((9·n) - 1) - (3·n - 1))/2

(3·n)·(((9·n) - 1) - (3·n - 1))/2 = (3·n)·(9·n - 1 - 3·n + 1)/2 = (3·n)·(6·n )/2 = 9·n²

Q₃.ₙ = (3·n)² = 9·n²

∴ P₃.ₙ - 3·Pₙ = P₃.ₙ - 3·Pₙ

Explanation: