The sum of the cubes of three consecutive nat | vedclass

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

Question

(C) Let the three consecutive natural numbers be $(n-1), n, (n+1)$ where $n \geq 2$. The sum of the cubes of these numbers is: $(n-1)^3 + n^3 + (n+1)^

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the three consecutive natural numbers be $(n-1), n, (n+1)$ where $n \geq 2$. The sum of the cubes of these numbers is: $(n-1)^3 + n^3 + (n+1)^3 = (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) = 3n^3 + 6n$. Factoring the expression,we get $3n(n^2 + 2)$. If $n$ is a multiple of $3$,then $3n$ is divisible by $9$. If $n$ is not a multiple of $3$,then $n^2 \equiv 1 \pmod{3}$,so $n^2 + 2 \equiv 1 + 2 \equiv 3 \equiv 0 \pmod{3}$. Thus,$n^2 + 2$ is divisible by $3$,making $3n(n^2 + 2)$ divisible by $3 	imes 3 = 9$. Therefore,the sum is always divisible by $9$.

The sum of the cubes of three consecutive natural numbers is divisible by

Similar Questions

For a positive integer $n$, $n^{3} + 2n$ is always divisible by

$a^n + b^n$ is divisible by which of the following if $n$ is any odd positive integer?

The remainder on dividing $1+3+3^{2}+3^{3}+\ldots+3^{2021}$ by $50$ is

Let $n \in \mathbb{N}$. Which one of the following is true?

For every natural number $n$,${3^{2n + 2}} - 8n - 9$ is divisible by

Vedclass Test Series

Exam Paper Generator

Online Exam Module