The expression n n -12 n -1- n - N is always divisible byA- 5B- 8C- 6D- 7

The correct option is C 6 nbsp;We know that sum (S) of squares of first n natural numbers is S=n(n+1)(2n+1)6 ⇒ n(n+1)(2n+1)=6S Since S is a natural number, n ...

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Standard XII

Mathematics

Sum of n Terms

The expressio...

Question

The expression $n (n + 1) (2 n + 1), n \in N$ is always divisible by

7

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Solution

The correct option is D $6$
We know that sum $(S)$ of squares of first $n$ natural numbers is $S = \frac{n (n + 1) (2 n + 1)}{6}$
$\Rightarrow n (n + 1) (2 n + 1) = 6 S$
Since $S$ is a natural number,
$n (n + 1) (2 n + 1)$ is divisible by $6.$

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The expression n(n+1)(2n+1),n∈N is always divisible by

The correct option is D 6 We know that sum (S) of squares of first n natural numbers is S=n(n+1)(2n+1)6 ⇒n(n+1)(2n+1)=6S Since S is a natural number, n(n+1)(2n+1) is divisible by 6.

The expression $n (n + 1) (2 n + 1), n \in N$ is always divisible by

The correct option is D $6$
We know that sum $(S)$ of squares of first $n$ natural numbers is $S = \frac{n (n + 1) (2 n + 1)}{6}$
$\Rightarrow n (n + 1) (2 n + 1) = 6 S$
Since $S$ is a natural number,
$n (n + 1) (2 n + 1)$ is divisible by $6.$