Question

prove that the square of every sequence is a term of this sequence 5,9,13.......​

Answer 1

Step-by-step explanation:

Sequence is 5,9,13,............

Common difference,  d=9−5=4

Thus, next term =13+4=17

tn=a+(n−1)d

⇒tn=5+(n−1)4=5+4n−4=1+4n

That is, from any term of this sequence, if we subtract 1, then we get a multiple of 4.

Now, 2012−1=2011

Since 2011 is not a multiple of 4, the term 2012 is not a term of this sequence.