The set {n(n+1)(2n+1) : n in mathbb{Z}} is a | vedclass

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

Question

(A) Let $f(n) = n(n+1)(2n+1)$.We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a

Vedclass · Accepted Answer

$\{6k : k \in \mathbb{Z}\}$

Vedclass · Answer

(A) Let $f(n) = n(n+1)(2n+1)$.We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a product of three consecutive integers.The product of three consecutive integers is always divisible by $3! = 6$.Therefore, $n(n+1)(n-1) = 6k_1$ and $n(n+1)(n+2) = 6k_2$ for some integers $k_1, k_2$.Thus, $f(n) = 6(k_1 + k_2) = 6k$ for some integer $k$.Hence, the set is a subset of $\{6k : k \in \mathbb{Z}\}$.

The set $\{n(n+1)(2n+1) : n \in \mathbb{Z}\}$ is a subset of:

Similar Questions

If $P, Q$ and $R$ are subsets of $A$,then $R \times (P \cup Q) = $

Write the following interval in set-builder form: $[6, 12]$

State whether the following statement is true or false. Justify your answer.
${2, 6, 10}$ and ${3, 7, 11}$ are disjoint sets.

Which of the following is a set? Justify your answer.
The collection of all boys in your class.

List all the elements of the following set:
$F = \{ x : x \text{ is a consonant in the English alphabet which precedes } k \}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module