find the sum of odd numbers from 1 to 99, inclusive?

Question

show work i have a F*king migrane and its getting late help PLS!!!

Answer 1

the number of odd numbers in total=(99-1)/2+1=50
The sum =(1+99)*50/2=2500

Answer 2

Cripes, anyone knows that in a sum like this you just pair up values from opposite ends. 99+1, 97+3, etc. All 100. There are 25 such pairs. So the answer is 2500 and you don't need more than one flat second to figure that out. If someone asked me to write a C++ program to print out the sum of all the odd numbers from 1 to 99 inclusive, with no further spec than that, I'd write them this: #include int main() { std::cout << 2500 << std::endl; return 0; } It's just too obvious and they'd deserve that result.

Answer 3

1 + 99 +
2 + 98 +
3 + 97 +
....
49 + 51 +
50

= (49 * 100) + 50 = 4950

Answer 4

Use the Gaussian method.

1+99 = 100
2+98 = 100
3+97 = 100
...
49+51 = 100
50 = 50
49*100+50 = 4950

OR if 99=(n-1), 100=n, then you can use [n(n-1)]/2 = (n^2)/2 - n/2

both equations work, you figure it out

Answer 5

"pair off" numbers

1+99=100
3+97=100
.
.
.
47+53=100
49+51=100

there are 25 such pairs.
25*100=2500

Hope your migraine is better

Answer 6

i agree with quilasheila.. don't have to put rude remarks in here if you want help.. it is not necessary..

Answer 7

Don't like your rude remarks.
Guess I will let you do it.
Came here to do math not to listen to you complain!