Math and Arithmetic
Algebra

What are all of the odd numbers from 1-99 inclusive added together?

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March 21, 2012 10:52PM

look at 1+3=4=2^2

1+3+5=9=3^2

1+3+5+7=16=4^2

The sum of the first n odd numbers is n^2!

Now how many are there?

look at 1-9 inclusive.. 1,3,5,7,9 there are 5

1-19-...there are 10 which is 20/2

so 1-99 should be 50 of them and the answer should be 50^2=2500

Think the young Euler's way:

-How many integer numbers are there from 1 to 100 inclusive? 100.

-How many of them are the odds and how many are the evens? 50 and 50.

So, from 1 to 100 inclusive there are 50 odds: 1, 3, 5, 7, 9, 11, ..., ..., ... , 95, 97 and 99.

Take them added together:

1 + 3 + 5 + 7 + 9 +11+13+...+93+95+97+99 (50 odds in total)

99+97+95+93+91+89+87+...+ 7+ 5 + 3 + 1 (50 odds in total)

Add the vertical pairs and their sums together:

(1+99)+(3+97)+(5+95)+(7+93)+(9+91)+...+(97+3)+(99+1).

-You can see that every pair added in each brachet gives 100 as a sum.

-How many pairs in brachets were added? 50 pairs.

So, since each pair gives 100 as a sum and there are 50 pairs in total, we take a total sum of 50x100=5000. But we took TWICE all the (50) odd numbers from 1 to 99 inclusive in order to add them in equal-summing pairs. So, we took TWICE their sum found out to be 5000. Then, their (single) sum must be 5000:2=2500.

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