in sequence 4,7,10,13....

PROVE the difference between squared of any two terms is a multiple of 3

since Tn = 1+3n

(Tn)^2 = 1+6n+9n^2
(Tk)^2 = 1+6k+9k^2

(Tn)^2 - (Tk)^2 = (1+6n+9n^2) - (1+6k+9k^2)
= 6n-6k + 9n^2-9k^2
= 3(2n-2k + 3n^2-3k^2)
which is a multiple of 3 for any k,n

thanks - very much.

To prove that the difference between the squares of any two terms in the sequence 4, 7, 10, 13, ... is a multiple of 3, we need to analyze the pattern and use mathematical reasoning.

First, let's observe the given sequence: 4, 7, 10, 13, ...

We notice that each term in the sequence is obtained by adding 3 to the previous term.

To prove that the difference between the squares of any two terms is a multiple of 3, let's consider two arbitrary terms in the sequence: the nth term (denoted as aₙ) and the mth term (denoted as aₘ), where n and m are any positive integers.

The nth term can be written as aₙ = 4 + 3(n-1) = 3n + 1
The mth term can be written as aₘ = 4 + 3(m-1) = 3m + 1

Now, let's find the difference between the squares of these two terms:
(aₙ)² - (aₘ)² = (3n + 1)² - (3m + 1)²
= 9n² + 6n + 1 - 9m² - 6m - 1
= 9n² - 9m² + 6n - 6m

Rearranging the terms, we get:
(aₙ)² - (aₘ)² = 9n² - 9m² + 6n - 6m
= 9(n² - m²) + 6(n - m)

We can factor out 3 from the expression:
(aₙ)² - (aₘ)² = 3(3(n² - m²) + 2(n - m))

Since n and m are positive integers, (n² - m²) and (n - m) are also integers.

Therefore, we have shown that the difference between the squares of any two terms in the given sequence is a multiple of 3.