Question

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

Answer 1

196

Explanation:

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

`S_n = n/2 [ 2a + (n - 1 )d ]`

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

`rArr S_14 = 14/2 [ (2xx1) + (13xx2) ] = 196`

Answer 2

Solution reworked and found to be 196

Explanation:

Series:`" "color(green)(1+3+5+7+9+...+27)`

`underline("Position count (n) | Term value | sum")`
`" 1 | 1 | 1"`
`" 2 | 3 | 4"`
`" 3 | 5 | 9"`
`" 4 | 7 | 16"`

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

`color(green)("So the number of terms is found by: "2n-1=27)`

`color(brown)(=> n = 28/2 = 14 " terms ")` which is an even number

`color(magenta)("So we have 7 pairs of numbers")`

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

There is a trick for solving these

Consider the above table

`" 1 3 5 7 "`
`" | "underline("| |" )" | "`
`" | | |"`
`" | "3+5=8" |"`
`" " underline("| |")"`
`" | "`
`" "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

`=>" Sum is " 14/2(1+27) = 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:

`" "[(n-1)/2("First"+"Last")] + "middle term"`

The `n-1` is mathematically excluding the middle term so you have to put it back with `"+ middle term"`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(red)("Play with this set up further and you realise that")`
`color(red)("that the sum is the mean value multiplied by n.")`
`color(red)("Still further investigation will reveal that the sum is "n^2)`

Answer 3

Development of solution formula being `n^2`

Explanation:

Let a term in the sequence `1+3+5+...+27` be `a_i`
Let the sum of this eries be `s`

Then `a_i=(2i-1)`

Consequently `s=sum_(i=1ton) a_i=sum_(i=1ton)(2i-1)`
sum_(I=1ton)
`" "s=2sum_(i=1ton)i-sum_(i=1ton)1`

But `sum_(i=1ton)1=n`

`s=2sum_(i=1ton)i-n`

but `sum_(i=1ton)i=nbari = n(1+n)/2`

`=>s=2n(1+n)/2-n`

`" "s" "= " "n+n^2-n" " = " "n^2`

`" "color(blue)(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`