Question

Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2.

Answer 1

Let the given statement P(n), be given as

P(n) : `sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2

We observe that

P(2) : `sum_(t = 1)^(2 - 1) t(t + 1) = sum_(t = 1)^1 t(t + 1)`

= 1.2

= `(1.2.3)/3`

= `(2.(2 - 1)(2 + 1))/3`

Thus, P(n) in true for n = 2.

Assume that P(n) is true for n = k ∈ N.

i.e., P(k) : `sum_(t = 1)^(k - 1) t(t + 1) = (k(k - 1)(k + 1))/3`

To prove that P(k + 1) is true

We have `sum_("t" = 1)^(("k" + 1 - 1)) "t"("t" + 1) = sum_("t" = 1)^"k" "t"("t" + 1)`

= `sum_("t" = 1)^(k - 1) t(t + 1) + k(k + 1)`

= `(k(k - 1)(k + 1))/3 + k(k + 1)`

= `k(k + 1)[(k - 1 + 3)/3]`

= `(k(k + 1)(k + 2))/3`

= `((k + 1)((k + 1) - 1)((k + 1) + 1))/3`

Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction, P(n) is true for all natural numbers n ≥ 2.