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

###### Wiki User

###### 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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