Question

Using mathematical induction, the numbers $a_n^{\prime }s$ are defined by, $a_0=1,a_{n+1}=3n^2+n+a_n(n\ge 0).$ Then $a_n=$

• A. $n^3+n^2+1$
• B. $n^3-n^2+1$
• C. $n^3-n^2$
• D. $n^3+n^2$

Answer 1

The correct option is B $n^3-n^2+1$

Given $a_{n+1}-a_n=3n^2+n$,
When $n=0,a_1-a_0=0$
When $n=1,a_2-a_1=4$
When $n=2,a_3-a_2=14$
.
.
When $n=n-1,a_n-a_{n-1}=3(n-1{)}^2+(n-1)$
Adding, we get $a_n-a_0=3\sum (n-1{)}^2+\sum (n-1)$
$\Rightarrow a_n-1=\frac{(n-1)n(2n-1)}{2}+\frac{(n-1)n}{2}=n^3-n^2$
$\Rightarrow a_n=n^3-n^2+1$
Let $P\left(n\right):a_n=n^3-n^2+1$ is true for $n.$
For $n=1\Rightarrow a_1=1,P(n+1):a_{n+1}=3n^2+n+n^3-n^2+1$
$\Rightarrow a_{n+1}=(n+1{)}^3-(n+1{)}^2+1$ is true for $n=n+1$ also.
So, $a_n=n^3-n^2+1$ is true.