prove this :$\sum_1^{100} A_i = \sum_0^{99} C_i $ In a class there are 100 student. We define $A_i$  as the number of friends of $ i^{th}$ student and $C_i$ as the number of students who has at least $i$ friends.
Prove that:
$\sum_1^{100} A_i = \sum_0^{99} C_i$
 A: Making the correction that the sum on the RHS should start from one instead of from zero:
Consider the number of times a specific student contributes to the sum on the RHS.
If a student has $i$ friends, then he will contribute one to the total in $C_1$, again in $C_2$, again in $C_3$, up until again in $C_i$.  I.e. it will contribute a total of $i$ to the overall sum (broken up over multiple parts of the summation).
The result follows immediately from that observation.

If that wasn't descriptive enough, consider it this way: let $\chi_{i,j} = \begin{cases} 1&\text{if student}~i~\text{has at least}~j~\text{friends}\\ 0&\text{otherwise}\end{cases}$
We have then the total sum is equal to:
$$\begin{array}{cccccc} \chi_{1,1}&+\chi_{1,2}&+\chi_{1,3}&+\chi_{1,4}&+\dots\\
\chi_{2,1}&+\chi_{2,2}&+\chi_{2,3}&+\dots\\
\chi_{3,1}&+\chi_{3,2}&+\dots\\
\vdots&&&\ddots\end{array}$$
Notice that $A_i$ is equal to the sum of the $i^{th}$ row whereas $C_j$ is equal to the sum of the $j^{th}$ column.  Stopping at $99$ for the sum for $C_j$ is fine since $C_{100}=0$ since noone can possibly be friends with $100$ people (there are only $99$ people for each person to be friends with).
So, the LHS summation is adding everything row by row whereas the RHS summation is adding everything column by column.
