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

3 Answers
Mar 14, 2016

Answer:

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 #

Mar 14, 2016

Answer:

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

Answer:

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#