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

3 Answers
Mar 14, 2016

196

Explanation:

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

S_n = n/2 [ 2a + (n - 1 )d ]Sn=n2[2a+(n1)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 S14=142[(2×1)+(13×2)]=196

Mar 14, 2016

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)

Mar 16, 2016

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