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In a certain test $a_i$ students gave wrong answers to at least i questions, where $i= 1,2,3......k$ No student gave more than k wrong answers. The total number of wrong answers is

I wasn't able to start solving this. I tried subtracting 2-1 for exactly 1 answer but I didn't understand what was happening. The solution provided uses the same logic from the below sum of Therefore, total number of wrong answers contributed by $2^{n−i} −2^{n−i−1}$ students who answered i questions wrong is $(2^{n−i}−2^{n−i−1})i$. I have no clue how this was derived.

There is a similar question on the site already but it is the other way around and doesn't resolve my doubt. To find number of questions when number of wrong answers is given

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    $\begingroup$ Did you mean to write $(2^{n - i} - 2^{n - i - 1})i$? This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jul 25 '20 at 10:01
  • $\begingroup$ @N.F.Taussig okay ill check it out thanks! $\endgroup$ – Shaurya Goyal Jul 25 '20 at 10:03
  • $\begingroup$ HINT: How many students gave precisely $i$ wrong answer? $\endgroup$ – FormulaWriter Jul 25 '20 at 10:14
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  • The number who gave at least $k$ wrong answers is $a_k$. Since no student gave more than $k$ wrong answers, the number who gave exactly $k$ wrong answers is $a_k$.

  • The number who gave at least $k-1$ wrong answers is $a_{k-1}$. Since $a_k$ students gave more than $k-1$ wrong answers, the number who gave exactly $k-1$ wrong answers is $a_{k-1}-a_k$.

  • The number who gave at least $k-2$ wrong answers is $a_{k-2}$. Since $a_{k-1}$ students gave more than $k-2$ wrong answers, the number who gave exactly $k-2$ wrong answers is $a_{k-2}-a_{k-1}$.

  • $\cdots$

  • The number who gave at least $1$ wrong answer is $a_{1}$. Since $a_2$ students gave more than $1$ wrong answer, the number who gave exactly $1$ wrong answers is $a_1-a_2$.

So the total number of wrong answers is $$1(a_1-a_2)+2(a_2-a_3) +\cdots+(k-1)(a_{k-1}-a_k)+k(a_k)$$ $$=a_1 +a_2(2-1)+a_3(3-2)+\cdots+a_k(k-(k-1))$$ $$=a_1+a_2+a_3+...+ a_k$$ $$=\sum\limits_1^k a_n$$

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  • $\begingroup$ I do not understand the second line. ak-1 is < ak. So the value come out on negative ? Could you please simplify it a little bit, or maybe just word it in a different way ? I tried this was ak=5 and ak-1= 4 but did not get it. $\endgroup$ – Shaurya Goyal Jul 26 '20 at 17:51
  • $\begingroup$ I got it. I was assumimg ak> ak-1 when the values would actually be converging. Thanks! $\endgroup$ – Shaurya Goyal Jul 26 '20 at 17:53

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