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

Among the statements:
$(S1):$ $2023^{2022} - 1999^{2022}$ is divisible by $8$.
$(S2):$ $13(13^{n}) - 11n - 13$ is divisible by $144$ for infinitely many $n \in N$.

If $a, b$ and $n$ are natural numbers, then $a^{2n-1} + b^{2n-1}$ is always divisible by:

Given below are two statements:
Statement $I$: $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by $7$.
Statement $II$: The integral part of $(7 + 4\sqrt{3})^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below:

The numbers $a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ when divided by $25$ leave the remainder:

The remainder when $7^n - 6n - 50$ $(n \in N)$ is divided by $36$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module