The numbers a_n are defined by a_0=1 and a_{n | vedclass

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(B) Given,$a_0=1$ and $a_{n+1}=3n^2+n+a_n$. We can find the first few terms: $a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$. $a_2 = 3(1)^2 + 1 + a_1 = 3 + 1

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$n^3-n^2+1$

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(B) Given,$a_0=1$ and $a_{n+1}=3n^2+n+a_n$. We can find the first few terms: $a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$. $a_2 = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$. $a_3 = 3(2)^2 + 2 + a_2 = 12 + 2 + 5 = 19$. Now,check the options: For $n=0$: $a_0 = 0^3 - 0^2 + 1 = 1$ (Matches). For $n=1$: $a_1 = 1^3 - 1^2 + 1 = 1$ (Matches). For $n=2$: $a_2 = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5$ (Matches). For $n=3$: $a_3 = 3^3 - 3^2 + 1 = 27 - 9 + 1 = 19$ (Matches). Thus,$a_n = n^3 - n^2 + 1$ is the correct expression.

The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

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