# Prove that the sum of 6 consecutive odd numbers is an even number?

Mar 6, 2018

#### Explanation:

Any two consecutive odd numbers add up to an even number.

Any number of even numbers when added result in an even number.

We can divide six consecutive odd numbers in three pairs of consecutive odd numbers.

The three pair of consecutive odd numbers add up to three even numbers.

The three even numbers add up to an even number.

Hence, six consecutive odd numbers add up to an even number.

Mar 6, 2018

Let first odd number be $= 2 n - 1$, where $n$ is any positive integer.

Six consecutive odd numbers are

$\left(2 n - 1\right) , \left(2 n + 1\right) , \left(2 n + 3\right) , \left(2 n + 5\right) , \left(2 n + 7\right) , \left(2 n + 9\right)$

Sum of these six consecutive odd numbers is

$\sum = \left(2 n - 1\right) + \left(2 n + 1\right) + \left(2 n + 3\right) + \left(2 n + 5\right) + \left(2 n + 7\right) + \left(2 n + 9\right)$

Adding by brute force method

$\sum = \left(6 \times 2 n\right) - 1 + 1 + 3 + 5 + 7 + 9$

We see that first term will always be even

$\implies \sum = \text{even number} + 24$

Since $24$ is even and sum of two even numbers is always even

$\therefore \sum = \text{even number}$

Hence Proved.

Mar 6, 2018

See below

#### Explanation:

An odd number has the form $2 n - 1$ for every $n \in \mathbb{N}$

Let be the first $2 n - 1$ we know that odd numbers are in arithmetic progresion with difference 2. So, the 6th will be $2 n + 9$

We know also that the sum of n consecutive numbers in a arithmetic progresion is

${S}_{n} = \frac{\left({a}_{1} + {a}_{n}\right) n}{2}$ where ${a}_{1}$ is the first and ${a}_{n}$ is the last one; $n$ is the number of sum elements. In our case

S_n=((a_1+a_n)n)/2=(2n-1+2n+9)/2·6=(4n+8)/2·6=12n+24

which is an even number for every $n \in \mathbb{N}$ because is divisible by 2 allways

Mar 6, 2018

$\text{We can actually say more:}$

$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$

$\text{Please see proof below.}$

#### Explanation:

$\text{We can actually say more:}$

$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$

$\text{Here's why. First, it is easy to see:}$

$\setminus q \quad \setminus q \quad \text{an odd number" + "an odd number" \ = \ "an even number}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{and}$

$\setminus q \quad \setminus q \quad \text{an even number" + "an even number" \ = \ "an even number} .$

$\text{Using these observations with the sum of any 6 odd numbers,}$
$\text{we see:}$

$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd}}_{6} \setminus =$

$\setminus q \quad \setminus \overbrace{{\text{odd"_1 +"odd"_2 }^{ "even"_1 } + \overbrace{ "odd"_3 +"odd"_4 }^{ "even"_2 } + \overbrace{ "odd"_5 +"odd"_6 }^{ "even}}_{3}} \setminus =$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus {\text{even"_1 + "even"_2 + "even}}_{3} \setminus =$

 \qquad \qquad \qquad \qquad \quad \ \ \overbrace{ "even"_1 + "even"_2 }^{"even"_4 } + "even"_3 \ =

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus {\text{even"_4 + "even}}_{3} \setminus =$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {\text{even}}_{5.}$

$\text{So we have shown:}$

$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$

$\text{So we conclude:}$

$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$