# How do you find the sum of the arithmetic series 1 + 3 + 5 + ... + 27?

Mar 14, 2016

196

#### Explanation:

The sum to n terms of an Arithmetic sequence is given by:

${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$

where a , is the 1st term , d the common difference and n , the number of terms to be summed.

Here a = 1 , d = 2 and n = 14

$\Rightarrow {S}_{14} = \frac{14}{2} \left[\left(2 \times 1\right) + \left(13 \times 2\right)\right] = 196$

Mar 14, 2016

Solution reworked and found to be 196

#### Explanation:

Series:$\text{ } \textcolor{g r e e n}{1 + 3 + 5 + 7 + 9 + \ldots + 27}$

$\underline{\text{Position count (n) | Term value | sum}}$
$\text{ 1 | 1 | 1}$
$\text{ 2 | 3 | 4}$
$\text{ 3 | 5 | 9}$
$\text{ 4 | 7 | 16}$

Notice that each term is calculated by $2 n - 1$

$\textcolor{g r e e n}{\text{So the number of terms is found by: } 2 n - 1 = 27}$

$\textcolor{b r o w n}{\implies n = \frac{28}{2} = 14 \text{ terms }}$ which is an even number

$\textcolor{m a \ge n t a}{\text{So we have 7 pairs of numbers}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There is a trick for solving these

Consider the above table

$\text{ 1 3 5 7 }$
$\text{ | "underline("| |" )" | }$
$\text{ | | |}$
$\text{ | "3+5=8" |}$
$\text{ " underline("| |")}$
$\text{ | }$
$\text{ } 7 + 1 = 8$

So for an even count of values the sum is

" "n/2("First"+"Last")

We have $n = 14$ terms which is even so we can apply this method

$\implies \text{ Sum is } \frac{14}{2} \left(1 + 27\right) = 196$

'~~~~~~~~~~~~~~~~~ Further comment ~~~~~~~~~~~~~~~~~
Suppose there had been an odd number of terms.

We could pair up our values as above but there would be a single unpaired value in the middle. In this case you would have:

$\text{ "[(n-1)/2("First"+"Last")] + "middle term}$

The $n - 1$ is mathematically excluding the middle term so you have to put it back with $\text{+ middle term}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{red}{\text{Play with this set up further and you realise that}}$
$\textcolor{red}{\text{that the sum is the mean value multiplied by n.}}$
$\textcolor{red}{\text{Still further investigation will reveal that the sum is } {n}^{2}}$

Mar 16, 2016

Development of solution formula being ${n}^{2}$

#### Explanation:

Let a term in the sequence $1 + 3 + 5 + \ldots + 27$ be ${a}_{i}$
Let the sum of this eries be $s$

Then ${a}_{i} = \left(2 i - 1\right)$

Consequently $s = {\sum}_{i = 1 \to n} {a}_{i} = {\sum}_{i = 1 \to n} \left(2 i - 1\right)$
sum_(I=1ton)
$\text{ } s = 2 {\sum}_{i = 1 \to n} i - {\sum}_{i = 1 \to n} 1$

But ${\sum}_{i = 1 \to n} 1 = n$

$s = 2 {\sum}_{i = 1 \to n} i - n$

but ${\sum}_{i = 1 \to n} i = n \overline{i} = n \frac{1 + n}{2}$

$\implies s = 2 n \frac{1 + n}{2} - n$

$\text{ "s" "= " "n+n^2-n" " = " } {n}^{2}$

$\text{ } \textcolor{b l u e}{s = {n}^{2}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Suppose the series did not start at 1 but was say: 15 to 47

You could calculate the sum from 1 to 47 and then subtract from it the sum of 1 to 13. So you would have ${47}^{2} - {13}^{2}$