## Sum of consecutive odd numbers

Let   $n$   be a positive integer and   $m$   be any integer with the same parity as   $n$
with   $m \; \geq \; n$

These odd integers are all positive.

for example,

If   n = 2,    m = 4

$m \times n = 8$

8   =   3 + 5

the product is equal to the sum of 2   (n = 2)   consecutive odd integers

If   n = 3,   m = 5

$m \times n = 15$

15   =   3 + 5 + 7

the product is equal to the sum of 3   (n = 3)   consecutive odd integers

If   n = 5,    m = 7

$m \times n = 35$

35   =   3 + 5 + 7 + 9 + 11

the product is equal to the sum of 5   (n = 5)   consecutive odd integers

Can you prove the product   $m \times n$   will always equal to the sum of   $n$   consecutive odd integers?

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Proof   –

The sum of the first   $n$   odd integers is   $n^2$ :

$1 \; + \; 3 \; + \; 5 \; + \; ... \; + \; (2 n-1) \; = \; n^2$

From this, we have that the sum of any sequence of consecutive odd integers is :

$(2m+1) + (2m+3) + ... + (2n-1)$

$= \; (1 + 3 + 5 + ... + (2n-1)) \; - \; (1 + 3 + 5 + ... + (2m-1))$

$= \; n^2 \; - \; m^2$

$n^2 \; - \; m^2 \; = \; (n+m) (n-m)$

So an integer   $k$   can be written as the sum of consecutive odd integers if it can be expressed as

$k \; = \; a b$

where

$a \; = \; n+m$        and        $b \; = \; n-m$

$a+b \; = \; 2 n$        or        $n \; = \; (a+b)/2$

$a-b \; = \; 2 m$        or        $m \; = \; (a-b)/2$

$a$   is odd,    $b$   is odd :

If   $k \; = \; 3 \times 5 \; = \; 15$

$n \; = \; (5+3)/2 = 4$        $m \; = \; (5-3)/2 = 1$

The first odd integer in the sequence is:   2(1) + 1 = 3

The last odd integer in the sequence is   2(4) – 1 = 7

We have the solution   $15 \; = \; 3+5+7$

$a$   is even,    $b$   is even :

E.g.   $12 \times 6 \; = \; 72$

a = 12,    b = 6

n = (a+b)/2         n = (12+6)/2 = 9
m = (a-b)/2         m = (12-6)/2 = 3

1st odd integer is :      2(3+1) – 1 = 7
last odd integer is :      2(9) – 1 = 17

We have the solution   7 + 9 + 11 + 13 + 15 + 17   =   72