Question

In a certain test with `k` questions, `a_1` students gave at least one wrong answer, `a_2` students gave at least `2` wrong answers, etc. What was the total number of wrong answers?

Answer 1

`sum_(i=1)^k a_i`

Explanation:

This is like the total area of a histogram with the `x` and `y` axes flipped. The area is unchanged by flipping the axes, but the perspective is.

Answer 2

The total number of wrong answer is given by
`a_1+a_2+...+a_k=sum_(i=1)^ka_i`

Explanation:

Note first that if a student gave at least `m` wrong answers, and `m > n`, then the student gave at least `n` wrong answers as well. So, for example, if a student gave `k` wrong answers, then that student also gave at least `k-1` wrong answers and at least `k-2` answers and so on.

With the above in mind, we can tell that the number of students who got exactly `i` answers wrong is `a_i-a_(i+1)` for `i < k` and `a_k` for `i=k`.

With that, we can find the total number of wrong answers by taking the sum over `i` of the number of students who got exactly `i` answers wrong multiplied by `i`. That is,

`"Total wrong" = [sum_(i=1)^(k-1)i(a_i-a_(i+1))]+ka_k`

`=a_1-a_2+2a_2-2a_3+...+(k-1)a_(k-1)-(k-1)a_k+ka_k`

`=a_1+(2-1)a_2+(3-2)a_3+...+(k-(k-1))a_k`

`=a_1+a_2+...+a_k`

`=sum_(i=1)^ka_i`

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We could also arrive at the result more quickly by noting that if a student got exactly `m` answers wrong, then they increase the value of `a_i` by `1` for each `i=1,2,...,m`. Then, if we sum all of the `a_i`'s, that student will be counted in `m` of them, meaning their contribution is equal to the number of their wrong answers. As this is true for each student, the total wrong answers given by all of them must be `sum_(i=1)^ka_i`