Question

1,4,7,10 are the sequences prove that the square of any term of this sequences is also a term of this sequences

Answer 1

To solve this, first we need to find a general term for the series.

Looking at the equation, we see that the numbers are getting added by 3, so a fitting general term would be 1+3n. Substituting n as any natural number, we can get all the terms of the series.

Now since we need to prove that the squares of all the terms of the series also belong in the series, we will square the general term.

$(1+3n)^{2}= 1+ 9n^{2}+ 6n$

$1+ 9n^{2}+ 6n = 1+ 3(3n^{2}+ 2n) = 1 + 3k$

(where k= $3n^{2} + 2n$)

Since the square of the general term: 1+3k can be written in the same form as the normal general term, therefore it belongs to the series as well.