## Consecutive Counting Groups

Group the consecutive counting numbers as follows: (1), (2,3), (4,5,6), (7,8,9,10), . .  Notice that there is one number in the first group, two numbers in the second group, three in the third, etc.  What is the sum of the numbers in the 99th group?
Source: mathcontest.olemiss.edu 1/18/2010

SOLUTION
This problem is similar to the problem “199 Numbers 3/12/2007″.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 1 2 3 4 5 6 7

The above table displays the groups 1 through 7. Given a group number, two important facts we want to establish are: 1) what is the starting number of the group and 2) what is the ending number of the group.

What is the starting number of a group?
Take group 5 for example. There are 4 groups in front of it and these 4 groups have a total of $1+2+3+4=10$ numbers starting from 1 through 10. So, group 5 must start from 11.
Similarly, group 7 has 6 groups in front of it. These 6 groups have a total of $1+2+3+4+5+6=21$ numbers starting from 1 through 21. Therefore, group 7 must start from 22.
In general, the starting number of group n is given by
$1+\sum_{i=1}^{n-1}i$

What is the ending number of a group?
Take group 7 for example. Its starting number is 22. Its ending number is $22+7-1=28$.
In general, the ending number of group n is given by
$1+\sum_{i=1}^{n-1}i+n-1=\sum_{i=1}^{n}i$

Armed with these two formulas we are now ready to solve the problem of group 99.

What is the starting number of group 99?
$1+\sum_{i=1}^{98}i=1+\frac{\left (98+1\right )98}{2}$
$=1+\left (99\right )49$
$=1+4851$
$=4852$

What is the ending number of group 99?
$\sum_{i=1}^{99}i=\frac{\left (99+1\right )99}{2}$
$=\left (50\right )99$
$=4950$

We calculate the sum of the 99 numbers of group 99 as follows
$\frac{\left (4950+4852\right )\left (4950-4852+1\right )}{2}$
$=\frac{\left (9802\right )\left (99\right )}{2}$
$=4901\left (99\right )$
$=485199$