Find the recursive formula for $a_{n}={b^n}$ for $n>2$

Question

How can I solve it, and can anybody help me to find recursive formula for $a_{n}=3n^3$ for $n\geq 0$.

Answer 1

Set $b_n=a_{n+1}-a_n=3(n+1)^3-3n^3=9n^2+9n+3$

Set $c_n=b_{n+1}-b_n=9(n+1)^2+9(n+1)+3-(9n^2+9n+3)=18n+18$

So $c_{n+1}=c_n+18$ and $c_0=18$

Then $b_{n+2}-b_{n+1}=b_{n+1}-b_n+18$

then $a_{n+3}-a_{n+2}-(a_{n+2}-a_{n+1})=a_{n+2}-a_{n+1}-(a_{n+1}-a_n)+18$

So $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n+18$, with $a_0=0, a_1=3, a_2=24$

Answer 2

My suggestion would be to try to consider $a_n=3((n-1)+1)^3$ and see what you can get out of it.

Answer 3

We have $a_{n+1}=3(n+1)^3=3(n^3+3n^2+3n+1)=a_n+9n^2+9n+3$. And $a_0=0$.