Answer:
The nth term of the sequence is:
[tex]a_{n}=2n^{2}-1[/tex]
Step-by-step explanation:
First, let's find the difference between the consecutive terms.
The difference between the 2nd and 1st term is: [tex]6[/tex]
The difference between the 3rd and 2nd term is: [tex]10[/tex]
The difference between the 4th and 3rd term is: [tex]14[/tex]
The difference between the 5th and 4th term is: [tex]18[/tex]
We can clearly see that the difference increases by 4, for each term.
So, the difference themselves have a sequence in them, with the initial term being: [tex]6[/tex], and the difference being: [tex]4[/tex].
According to the formula for nth term:
[tex]a_{n}=a_{1}+(n-1)d[/tex]
Substitute the values:
[tex]a_{n}=6+(n-1)\cdot 4\\\\\text{Simplify}:\\a_{n}=6+4n-4\\a_{n}=2+4n[/tex]
We will substitute [tex]n[/tex] for [tex]n-1[/tex], so that for the second term of the entire sequence, the difference added becomes [tex]6[/tex]
[tex]d=4+2(n-1)~~~~~d=4+2(2-1)~~~~~d=6[/tex]
Simplify the equation now:
[tex]d=4+2(n-1)\\d=4+2n-2\\d=2+2n[/tex]
So, the difference itself is governed by the sequence: [tex]d_{n}=2+2n[/tex].
Now, for the entire sequence itself, the values are:
[tex]a_{1}=1[/tex]
[tex]d=2+2n[/tex]
Substitute them into the formula for nth term:
[tex]a_{n}=1+(n-1)(2+2n)[/tex]
Complete the rest of the simplification:
[tex]\text{Applying the distributive property:}\\a_{n}=1+ n(2+2n)-1(2+2n)\\\\a_{n}=1+2n+2n^{2}-2-2n\\\\\text{Simplify:}\\a_{n}=2n^{2}-1[/tex]
