word problem

avinashsidha wrote:For any positive integer n,the sum of first n positive integers
equals n(n+1)/2 . What s the sum of all even integers between 99 and 301

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 =20,200

Cheers,
Brent

Alternatively, if we want to evaluate 2(50+51+52+...+149+150), we can evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200

Cheers,
Brent

SoftwareDrone wrote:Before I saw the elegant answers, I tried this off the top of my head:
If you can figure out the sum of consecutive integers by:

100 + 99 + 98 + ... + 3 + 2 + 1
1 + 2 + 3 + ... + 98 + 99 + 100
= (101 * 99) / 2

I figured that I could do this:
100 + 102 + 104 + ...
300 + 298 + 296 + ...
= 400 * 101 = 40400 / 2 = 20200.
I think this will work, right?

Perfect solution.
The only difficult part with that approach is recognizing that there are 101 even integers between 99 and 301. Once you figure that out (as you did), everything falls into place.
Nice work.

Cheers,
Brent