sum of consecutive natural numbers

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by hemangsarkar, Aug 23, 2015, 6:42 PM

Q) Find all natural numbers $n>2$ such that $n$ can be written as a sum of consecutive natural numbers.

Solution:

If $p>2$ and $p$ is odd, then $p = 2k+1 = (k) + (k+1)$ .

Let $p = 2k$ be even.

let $(m+1) + (m+2) + .... + (n) = 2k$

where $m \geq 0$

$\frac{n(n+1)}{2} - \frac{m(m+1)}{2} = 2k$

$n^2 + n - m - m^2 = 4k$

$(n-m)(n+m) + (n-m) = 4k$

$(n-m)(n+m+1) = 4k$

now, $(n+m+1) - (n-m) = 2m+1$ .

if $(n+m+1)$ is odd, then $(n-m)$ is even, if $(n+m+1)$ is even, then $(n-m)$ is odd.

We can't get any solution if the right hand side is purely even.

Hence no solution when $k$ is a power of $2$ .

Else we can get $(n-m)(n+m+1) = 4(s)(2^t)= (s)(2^{t+2})$

where $s$ is odd and $t \geq 0$ . Equating factors, we get

$n-m = s$

$n+m+1 = 2^{t+2}$

$n = \frac{2^{t+2}+s-1}{2}$ and $m = \frac{2^{t+2}-s-1}{2}$

provided $2^{t+2} \geq s + 1$

or

$n-m = 2^{t+2}$

$n+m+1 = s$

$n = \frac{2^{t+2}+s-1}{2}$ and $m = \frac{s-2^{t+2}-1}{2}$

provided $s \geq 2^{t+2} + 1$

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sum of consecutive natural numbers

by hemangsarkar, Aug 23, 2015, 6:42 PM

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