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?

2 Answers
Jun 12, 2016

Answer:

#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.

enter image source here

Jun 12, 2016

Answer:

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#


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#