Sum of consecutive odd integers

Number Theory Level 2

The sum of $n$ consecutive odd integers is a multiple of $n$ .

For what positive integers $n$ is this statement true?

All positive integers

n

n

odd only

n

even only

n

multiple of 3 only

1 solution

Nelson Mandela
Aug 9, 2015

Sum of n consecutive integers is n^2.

1+3+5+....+2n-1.(n terms).

It is in AP .

Sum = $\frac { n }{ 2 } (2(1)+(n-1)2)\quad =\quad \frac { n }{ 2 } (2+2n-2)\quad =\quad \frac { 2{ n }^{ 2 } }{ 2 } \quad =\quad { n }^{ 2 }$ .

So, every positive integer will satisfy the condition that n^2 is a multiple of n.

When we take the general case,

(2n+1)+(2n+3)+(2n+5)+........(2n+2n-1).{n terms}.

Sum = \frac { n }{ 2 } (2(2n+1)+(n-1)2)\quad =\quad \frac { n }{ 2 } (4n+2+2n-2)\quad =\quad 3{ n }^{ 2 }.

Which is also a multiple of n for all positive integers.

Thanks to @Chung Kevin for guiding me.

You have shown it for a very special case. Is it true of any $n$ consecutive odd integers?

Chung Kevin - 5 years, 10 months ago

Please check the solution now.

Nelson Mandela - 5 years, 10 months ago

Hm, in your new sequence, you just have the integers from $2n+1$ to $2n+2n-1$ . You would want $2k+1$ to $2k+2n-1$ instead.

I was thinking of it as "We will be adding $2k$ a total of $n$ times, hence it is clearly still a multiple of $n$ . In

Chung Kevin - 5 years, 10 months ago