Hello,

The formulation of the question says:

Show, using congruences and disjunction of cases, which,for all natural n є Ninteger, the integern (n+1) (2n+1) is a multiple of 6.

Simultaneously, themultiple of 3 is a multiple of 2 (number), then it is divisible by 6

3 cases are possible by applying the congruence module 3:

If $n=3k \Longrightarrow n 3k\left ( n+1 \right )$$\left ( 2n+1 \right )=3k'$

If $n=3k+1 \Longrightarrow 2n+1=6k+3$ $3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''$

If $n=3k+2 \Longrightarrow n+1=3k+3[/tex] and $$3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''$ Thank you very much for your attention  Quote by inverse Hello, The formulation of the question says: Show, using congruences and disjunction of cases, which,for all natural n є Ninteger, the integern (n+1) (2n+1) is a multiple of 6. Simultaneously, themultiple of 3 is a multiple of 2 (number), then it is divisible by 6 3 cases are possible by applying the congruence module 3: If $n=3k \Longrightarrow n 3k\left ( n+1 \right )$$\left ( 2n+1 \right )=3k'$ If $n=3k+1 \Longrightarrow 2n+1=6k+3$ $3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''$ If $n=3k+2 \Longrightarrow n+1=3k+3$$ and $$3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''$ Thank you very much for your attention You should be more careful before you send over a post, and that's what the "preview post" button exists for. You want to know for what positive integers n, the number $\,(n+1)(2n+1)$ is a multiple of 6. Your division in cases is fine, but for example: [tex]n=3k\Longrightarrow (n+1)(2n+1)=(3k+1)(6k+1)$$ which is never a multiple of 6 (I just can't understand what you did in this case)

Simmilarly for the next case, $$n=3k+1\Longrightarrow (n+1)(2n+1)=(3k+2)(6k+3)$$ which is multiple of 6 iff $\,k\,$ is even (why?).

Now try to do correctly the last case by yourself.

DonAntonio
 I cannot understand what are you asking me, can you ask the question again? thanks

Blog Entries: 2