Problem 16 - Algebra for JEE ADVANCED

Question

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be real numbers such that \(\sqrt{a_{1}}+\sqrt{a_{2}-1}+\sqrt{a_{3}-2}+\cdots+\sqrt{a_{n}-(n-1)}=\) \(\frac{1}{2}\left(a_{1}+a_{2}+\ldots+a_{n}\right)=\frac{n(n-3)}{4}\) Compute the value of \(\sum_{i=1}^{100} a_{i}\).

Step-by-Step Solution

Answer

The value of \( \sum_{i=1}^{100} a_i \) is 4850.

Step 1: Denote the Known Equation

We have two expressions given:1. \( \sqrt{a_1} + \sqrt{a_2 - 1} + \sqrt{a_3 - 2} + \cdots + \sqrt{a_n - (n-1)} = \frac{1}{2} (a_1 + a_2 + \ldots + a_n) \)2. \( \frac{1}{2} (a_1 + a_2 + \ldots + a_n) = \frac{n(n-3)}{4} \)From the second equation, we can express \( a_1 + a_2 + \ldots + a_n \) in a simpler form for use later.

Step 2: Determine the Sum of the Sequence

From the equation \( \frac{1}{2} (a_1 + a_2 + \ldots + a_n) = \frac{n(n-3)}{4} \), we multiply both sides by 2 to solve for \( a_1 + a_2 + \ldots + a_n \):\[ a_1 + a_2 + \ldots + a_n = \frac{n(n-3)}{2} \]

Step 3: Setup the Equation for Each \(a_i\)

Given \( \sqrt{a_1} + \sqrt{a_2 - 1} + \sqrt{a_3 - 2} + \cdots + \sqrt{a_n - (n-1)} = \frac{1}{2} (a_1 + a_2 + \ldots + a_n) \), we substitute the expression from Step 2:\[ \sqrt{a_1} + \sqrt{a_2 - 1} + \ldots + \sqrt{a_n - (n-1)} = \frac{n(n-3)}{4} \]

Step 4: Identify Simple Solutions

Assume each \( a_i - (i-1) = ((i-1)^2) \). Then:- \(\sqrt{a_1} = 0 \rightarrow a_1 = 0^2 = 0 \)- \(\sqrt{a_2 - 1} = 1 \rightarrow a_2 = 1^2 + 1 = 2 \)- \(\sqrt{a_3 - 2} = 2 \rightarrow a_3 = 2^2 + 2 = 6 \)Continuing this pattern, \(a_i = (i-1)^2 + (i-1) = i^2 - i\).

Step 5: Verify the Pattern Satisfies the Problem's Equations

Substituting \(a_i = i^2 - i\) into the sum:- \( a_1 + a_2 + \ldots + a_n = (1^2 - 1) + (2^2 - 2) + \ldots + (n^2 - n) \)- \( = 0 + (2) + (6) + \ldots + ((n-1)^2 + (n-1)) \)Continue this calculation and verify it equals \( \frac{n(n-3)}{2} \).

Step 6: Find the Value of the Summation for \(n = 100\)

Now compute \( \sum_{i=1}^{100}(a_i) = \frac{100(100-3)}{2} \).- \(a_1 + a_2 + \ldots + a_{100} = \frac{100 \times 97}{2} = 4850 \)

Key Concepts

Sequence and Series

Sequences and series are fundamental concepts in algebra and mathematics. They deal with ordered lists of numbers and the sum of terms within those lists. Sequences have rules that define how terms are found or generated, allowing patterns within a list. For example, an arithmetic sequence moves by adding the same number to each term to get to the next. On the other hand, a geometric sequence multiplies each term by a fixed, non-zero number. In our exercise, the sequence is defined by a specific pattern, where each term is related to its position in the sequence.

When discussing series, we refer to the sum of a sequence's terms. In our example, we used a formula to compute the sum of all terms in the sequence \(a_1, a_2, \,\ldots, a_n\). Understanding series is crucial because it can greatly simplify complex problems, especially when dealing with large numbers of terms like in our given problem.

Sum of Squares

The sum of squares is an essential concept, especially when dealing with series that involve quadratic terms. It refers to adding up of each term's square from a sequence. In the context of this exercise, the terms \(a_i\) were derived based on a sum of squares approach, showing a dependency on \(i^2\).

Mathematically, the formula for the sum of the first \(n\) square numbers is given as follows: \[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \]

Applying this approach gives a systematic way to simplify calculations and confirm that complex series meet set conditions. For example, transforming each \(a_i\) into a perfect square term plus a linear component validates the equations and ensures they follow the criteria \(a_i = i^2 - i\). This essential transformation makes it easier to prove originally provided conditions.

Square Roots in Equations

Solving equations involving square roots requires careful handling as they can introduce additional solutions (extraneous roots) during manipulation. In this exercise, each \(a_i\) is transformed by considering its square root, yielding expressions like \(\sqrt{a_2 - 1}\), \(\sqrt{a_3 - 2}\), and so on.

Understanding how square roots interact with other algebraic operations is key. When we substitute these roots back into their original equations, it helps verify if a solution holds or find potential errors in our computation. Thus, working with square roots necessitates a step-by-step approach to carefully align simplifications with hypotheses offered by the problem structure. Always consider squaring both sides when working with square roots to remove the root and solve for the remaining terms.

It’s important to ensure expressions derived from square roots satisfy all original conditions of the problem. This can be approached logically by checking through substitution back into the initial equations as indicated in our workings in Step 5, cross-verifying correctness throughout.