Question #e35c4

1 Answer
YoshiMan007
Oct 21, 2017

#2500#

Explanation:

#1+3+5+...+97+99#

Notice the following: the first and last term add to be:
#1+99=100#

The second and second-to-last:
#3+97=100#

Third and third-to-last:
#5+95=100#

This pattern continues until #49+51=100#

Now, all we need to do is figure out how many odd numbers there are from #1# to #49#, inclusive.

Well there are #5# odd numbers in every multiple of #10# (#1,3,5,7,9# or #11,13,15,17,19#). Since there are #5# multiples of #10# between #1# and #49# (the digits, the tens, the twenties, the thirties, and the forties), there are #5*5=25# odd numbers.

#25*100=2500#.

Just as a side note, it can be shown that the sum of the odd numbers from #1# to #2n+1# is #n^2#.

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