A=1+3+5+7................. 1999 ?

This series has #(1999-1)/2+1 = 1000# terms.

The average term is the same as the average of the first and last elements, namely #(1+1999)/2 = 1000#

So the sum is #1000*1000 = 1000000#

We are working with an arithmetic series.

#1+3+5+7+....... 1999#

There are two formulae which we can use to find the sum:

#S_n = n/2[2a+(n-1)d]" or " n/2(a+l)#

We have the first term, #" "a = 1#
We have the common difference #" "d =2#
We have the last term #" "l = 1999#

The only value that is not given directly is the number of terms, #n#

but we can find it:

From #1# to #2000# there are #2000# numbers.

However, we are only using the odd numbers, which is half of the numbers.

So from #1# to #1999# there are #1000# odd numbers

Now we have the number of terms: #n = 1000#

The second formula is much easier and quicker to use:

#S_1000 = 1000/2(1+1999)#

#S_1000 =500xx2000#

#S_1000 = 1,000,000#

If there is an even count of numbers then use condition A
If there is an odd count of numbers then use condition B

#color(blue)("Test which condition to use.")#

Let the test value be #t#

#("last number "xx1/2) + 1/2 ->" t"#

#color(brown)("If the test value is even then use condition A")#

#color(brown)("If the test value is odd then use condition B")#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the repeated value" -> 2t)#

You can use #2t# or #1+"last value"#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the count of repeats")#

#color(brown)("For even counts")#
#1+3+5+7 -> 2" repeat "->1/2(7/2+1/2)=2#
#1+3+5+7+9+11 -> 3" repeats "->1/2(11/2+1/2)=3#

#color(brown)("For odd counts")#
#1+3+5 ->1" repeat "->1/2(5/2-1/2)=1#
#1+3+5+7+9 ->2" repeat "->1/2(9/2-1/2)=2#

For odd count you have to add the centre value to the sum of the repeats.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("Answering your question")#

Even or odd?#->(1999xx1/2) + 1/2 ->1000 ->" even"#

Thus condition A

Repeated value #->1+1999=2000#

Count of repeats#->1/2(1999/2+1/2)" "=" "500#

Thus #1+3+5+7+...+1999 = 500xx2000 = 1000,000#

Let the #i^("th")# term be #a_i#

Then we have #sum_(i=1ton)a_i ->1+3+5+7+...+1999#

So the actual values are:

#i->color(white)(.)1color(white)(+)2color(white)(+)3color(white)(+)4color(white)(+)5" "..." "n#
#a_i-> 1color(white)(+)3color(white)(+)5color(white)(+)7color(white)(+)9" "..." "1999#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We need to determine a way to always have the correct odd value in relation to its position in the sequence.

To always have an even number we double the place number. Logically to always have an odd number we step back by 1
#=>#even -1 = odd

So lets test #a_i = 2i-1#

#i->color(white)(...)1color(white)(+)2color(white)(+)3color(white)(+)4color(white)(+)5" "..." "n#
#ul(2i->color(white)(.) 2color(white)(+)4color(white)(+)6color(white)(+)8color(white)(+)10" "..." "2n)#
#-1->1color(white)(+)3color(white)(+)5color(white)(+)7color(white)(+)9" "..." "2n-1 #
This works.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let the last term be #L=2n-1=1999#

Thus #2n=1999+1=2000#

#" "=>n=1000 larr" count"#

The mean value #=(1+L)/2#

#color(green)("The above is the link to the diagrams in my other approach")#

multiply by the count to obtain the sum:

#" "->(n(1+L))/2 #

but #n=1000 and L=1999#

#" "->(1000(1+1999))/2 = 1000^2#

The sum is #" " 1,000,000#

Calling #n = (1999-1)/2=999#

#S_n = sum_(k=0)^n (2k+1) = 2sum_(k=0)^nk + n+1 = 2((n+1)n)/2+n+1 = (n+1)n+n+1 = (n+2)n+1 = (999+2)999+1=1000000#