Using mathematical induction,the numbers a_n | vedclass

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(B) Given the recurrence relation: $a_0=1$ and $a_{n+1}=3n^2+n+a_n$. We calculate the first few terms: For $n=0$: $a_1 = 3(0)^2+0+a_0 = 1$. For $n=1$:

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$n^3-n^2+1$

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(B) Given the recurrence relation: $a_0=1$ and $a_{n+1}=3n^2+n+a_n$. We calculate the first few terms: For $n=0$: $a_1 = 3(0)^2+0+a_0 = 1$. For $n=1$: $a_2 = 3(1)^2+1+a_1 = 3+1+1 = 5$. For $n=2$: $a_3 = 3(2)^2+2+a_2 = 12+2+5 = 19$. Now,we test the options for $n=3$: Option $A$: $3^3+3^2+1 = 27+9+1 = 37$. Option $B$: $3^3-3^2+1 = 27-9+1 = 19$. Option $C$: $3^3-3^2 = 27-9 = 18$. Option $D$: $3^3+3^2 = 27+9 = 36$. Since $a_3=19$,Option $B$ is the correct formula.

Using mathematical induction,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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