Answer
(a) See explanations.
(b) See explanations.
(c) $a_n=n^2+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$
(d) $a_n=2^{n}$
$b_n=b_{n-1}+2b_{n-2}$ where $n\gt2$ with $b_1=2, b_2=4$
Work Step by Step
(a) Given $a_n=n^2$, we have: $a_1=1^2=1$, $a_2=2^2=4$, $a_3=3^2=9$, and $a_4=4^2=16$.
(b) Given $a_n=n^2+(n-1)(n-2)(n-3)(n-4)$, we have: $a_1=1^2+0=1$, $a_2=2^2+0=4$, $a_3=3^2+0=9$, and $a_4=4^2+0=16$.
(c) Follow the results from part (a) and (b), we can define $a_n=n^2+(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ which will satisfy the conditions.
(d) We can define $a_n=2^{n}$ which will give $2,4,8,16,...$
We can also define $b_n=b_{n-1}+2b_{n-2}$ where $n\gt2$ with $b_1=2, b_2=4$ and this series will also give $2,4,8,16,...$.
