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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$n^3-n^2+1$

Vedclass · Answer

(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:

Similar Questions

The $n^{\text{th}}$ term of the series $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ is:

$1+\sin x+\sin ^2 x+\sin ^3 x+\ldots+\infty=4+2 \sqrt{3}$ and $0 < x < \pi, x \neq \frac{\pi}{2}$,then $x=$

How many terms are there in the sequence $1, 3, 6, 10, 15, 21, \dots, 5050$?

$\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1 = \dots$

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module