If n is a positive integer,then n^{3}+2n is d | vedclass

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

Question

(D) Let $P(n) = n^{3} + 2n$. For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$. For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$. For $n = 3$,$P(3) = 3^{3}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $P(n) = n^{3} + 2n$. For $n = 1$,$P(1) = 1^{3} + 2(1) = 1 + 2 = 3$. For $n = 2$,$P(2) = 2^{3} + 2(2) = 8 + 4 = 12$. For $n = 3$,$P(3) = 3^{3} + 2(3) = 27 + 6 = 33$. We observe that $3, 12, 33$ are all divisible by $3$. Alternatively,$n^{3} + 2n = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$. Since $(n-1)n(n+1)$ is the product of three consecutive integers,it is always divisible by $3! = 6$. However,the expression $n^{3} + 2n$ is always divisible by $3$ for any positive integer $n$.

If $n$ is a positive integer,then $n^{3}+2n$ is divisible by

Similar Questions

Using mathematical induction,the numbers $a_n$ are defined by:
$a_0 = 1, a_{n+1} = 3n^2 + n + a_n, (n \geq 0)$.
Then,$a_n$ is equal to:

The values of the natural numbers $n$ for which the inequality $2^n > 2n + 1$ is valid are:

Prove that $2^n > n$ for all positive integers $n$.

Prove the following by using the principle of mathematical induction for all $n \in N:$
$1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1$

For all $n \in N$,$2^{2n+1} + 3^{2n+1}$ is divisible by

Vedclass Test Series

Exam Paper Generator

Online Exam Module