We will count number of squares
first term is 4
so, [tex] a_1=4 [/tex]
second term is 7
so, [tex] a_2=7 [/tex]
third term is 10
so, [tex] a_3=10 [/tex]
We can see that this sequence is in arithematic sequence
we can use nth term formula
[tex] a_n=a_1+(n-1)d [/tex]
where
an is nth terms
a1 is first term
d is common difference
n is total number of terms
[tex] d=7-4=3 [/tex]
now, we can plug values
and we get
[tex] a_n=4+(n-1)*3 [/tex]
[tex] a_n=4+3n-3 [/tex]
[tex] a_n=1+3n [/tex]
[tex] a_n=3n+1 [/tex].................Answer
