In a certain test, a_i students gave wrong answers to atleast i questions, where i = 1, 2, ..., k. N

Q: Wait, if I just add $a_1 + a_2 + \dots + a_k$, aren't I counting the same student multiple times? That feels like double-counting!

That is a brilliant observation! You are absolutely right—you *are* counting the same student multiple times. But here is the secret: that is exactly what you *want* to do. If a student made 3 mistakes, they *should* be counted 3 times—once for each mistake they made. The 'at least' definition naturally handles this multiplicity, ensuring that every single mistake is accounted for exactly once in the final sum.

Q: Why does the formula $n_i = a_i - a_{i+1}$ stop working at $i=k$? Is there a missing term?

Great catch! The formula doesn't stop working; it just hits a boundary condition. Since no student made more than $k$ mistakes, the number of students with 'at least $k+1$' mistakes is zero. So, $a_{k+1} = 0$. When you plug that into the formula, you get $n_k = a_k - 0 = a_k$. The logic holds perfectly!

Q: Is this just a counting trick, or is there a deeper mathematical principle here?

You have stumbled upon a beautiful, fundamental concept! This is a discrete version of a powerful theorem in probability and statistics: $E[X] = \sum_{x=1}^{\infty} P(X \ge x)$. You are essentially learning how to calculate an expectation by summing the 'tail' probabilities. This isn't just a trick; it is a foundational tool you will use in advanced mathematics for years to come.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are going to look at a problem that seems like a simple counting exercise but is actually a gateway to a profound mathematical truth. We are tasked with finding the total number of wrong answers given by a group of students, given the number of students who made 'at least $i$' mistakes.

Let $n_i$ be the number of students who made **exactly** $i$ mistakes. We assume $i$ ranges from $1$ to $k$, and since no one made more than $k$ mistakes, $n_i = 0$ for any $i > k$.

Let $a_i$ represent the number of students who made **at least** $i$ mistakes. Because a student who made $j$ mistakes (where $j \ge i$) is counted in $a_i$, we can express this relationship as:

$$a_i = n_i + n_{i+1} + \dots + n_k$$

### The Mathematical Bridge

Our goal is to find the total number of wrong answers, $W$. If we know how many students made exactly $i$ mistakes, the total is the sum of mistakes:

$$W = \sum_{i=1}^{k} i \cdot n_i = 1 \cdot n_1 + 2 \cdot n_2 + 3 \cdot n_3 + \dots + k \cdot n_k$$

We need to express $n_i$ in terms of $a_i$. By observing the definitions of $a_i$ and $a_{i+1}$, we see that:

$$a_i - a_{i+1} = (n_i + n_{i+1} + \dots + n_k) - (n_{i+1} + \dots + n_k) = n_i$$

Thus, for any $i < k$, we have $n_i = a_i - a_{i+1}$. For the final term, $n_k = a_k$, as there are no students with $k+1$ mistakes.

### The Telescoping Triumph

Now, let us substitute these expressions back into our total sum $W$:

$$W = 1(a_1 - a_2) + 2(a_2 - a_3) + 3(a_3 - a_4) + \dots + (k-1)(a_{k-1} - a_k) + k \cdot a_k$$

Expanding this expression, we obtain:

$$W = 1a_1 - 1a_2 + 2a_2 - 2a_3 + 3a_3 - 3a_4 + \dots + (k-1)a_{k-1} - (k-1)a_k + k \cdot a_k$$

Grouping the terms by $a_i$, we find the coefficient for each $a_i$:

$$W = 1a_1 + (-1+2)a_2 + (-2+3)a_3 + \dots + (-(k-1)+k)a_k$$

### Final Calculation

Every single coefficient simplifies to $1$. Therefore, the total number of wrong answers is simply the sum of the 'at least' counts:

**$$W = \sum_{i=1}^{k} a_i$$**

It is breathtakingly simple. We started with a complex, overlapping set of data and, through the power of algebraic manipulation, arrived at a result that is nothing more than the sum of our given values.

This is the beauty of mathematics—taking a seemingly chaotic problem and finding the underlying order. Keep this intuition close; it will serve you well in your journey through JEE Advanced and beyond.

Animated Solution for Mathematics - Permutations and Combinations: In a certain test, $a_{i}$ students gave wrong answers to atleast $i$ questions, where $i = 1, 2, \dots, k$ . No student gave more than $k$ wrong answers. The total number of wrong answers given is .........

Visualized Solution

Defining the Variable $n_{i}$

Relating $a_{i}$ to $n_{i}$

The Total Mistakes Formula

The $a_{i} - a_{i + 1}$ Relationship

Substituting and Expanding

Simplifying the Sum

The Way Forward

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Mathematical Bridge

The Telescoping Triumph

Final Calculation

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