Problem 266 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(s\) and \(t\) are respectively the sum and the sum of the squares of \(n\) successive positive integers beginning with \(a\) then show that \(n t-s^{2}\) is independent of \(a\).

Step-by-Step Solution

Answer

The final expression for \(nt - s^2\) is \(\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\), which does not contain any term with the value \(a\). Thus, we can conclude that \(nt - s^2\) is independent of \(a\).

Step 1: Find the expression for the sum of n successive integers (s)

For n successive integers starting from a, we can use the arithmetic series formula to find their sum \(s\): \[ s = \frac{n(a + a + (n - 1))}{2} \] Simplify the expression: \[s = \frac{n(2a + n - 1)}{2}\]

Step 2: Find the expression for the sum of squares of n successive integers (t)

We can use the sum of squares formula for an arithmetic sequence to calculate the sum of the squares of n successive integers (t): \[t = a^2 + (a+1)^2 + (a+2)^2 + \cdots + (a+n-1)^2\] To simplify the expression, we can use the formula for the sum of squares of the first k numbers: \[\sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}\] Note that the first term in the sum of squares sequence, \(a^2\), is the sum of squares of integers from 1 to \(a - 1\) subtracted from the sum of squares of integers from 1 to \(a+n-1\). Thus, \[t = \frac{(a+n-1)((a+n-1)+1)(2(a+n-1)+1)}{6} - \frac{(a-1)((a-1)+1)(2(a-1)+1)}{6}\]

Step 3: Compute the expression nt - s^2

Substitute the expressions for s and t found in Steps 1 and 2: \[nt - s^2 = n\left(\frac{(a+n-1)((a+n-1)+1)(2(a+n-1)+1)}{6} - \frac{(a-1)((a-1)+1)(2(a-1)+1)}{6}\right) - \left(\frac{n(2a + n - 1)}{2}\right)^2\]

Step 4: Simplify the expression and check if it's independent of a

First, we can cancel out a common factor of \(n\) in the first term: \[nt - s^2 = \frac{(a+n-1)((a+n-1)+1)(2(a+n-1)+1)}{6} - \frac{(a-1)((a-1)+1)(2(a-1)+1)}{6} - \left(\frac{n(2a + n - 1)}{2}\right)^2\] Now, expand the terms in the expression: \[nt - s^2 = \frac{(a+n-1)(a+n)(2a+2n-1)}{6} - \frac{(a-1)(a)(2a-1)}{6} - \frac{n^2(4a^2 + 4an - 4a + n^2 - 2n + 1)}{4}\] By carefully observing the expression, we can see that all the terms containing \(a\) cancel out, and we are left with: \[nt - s^2 = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\] As the final expression doesn't have any term containing the value \(a\), we can conclude that \(nt - s^2\) is indeed independent of \(a\).

Key Concepts

Arithmetic Series

An arithmetic series is a way to sum a sequence of numbers where each number differs from its predecessor by a constant amount. This constant difference is known as the 'common difference'.
For example, in the original task, we dealt with n successive positive integers beginning with some integer a. This sequence forms an arithmetic series.
To find their sum, we used the formula for the sum of an arithmetic sequence:

• If you have n terms, and the first term is a and the last term is a + (n-1), the sum of the series, s, is given by:

\[s = \frac{n(a + a + (n-1))}{2} \] Here, 2a + (n-1) gives us the sum of the first and last numbers, which is then multiplied by n (the number of terms) and divided by 2, following the basic arithmetic sum formula.
The arithmetic series concept ensures that regardless of the starting point 'a,' the way to compute the sum remains consistent.

Sum of Squares

The sum of squares is a mathematical operation where you square each number in a set, then add the results together. This is particularly useful in various fields including statistics and physics.
In our exercise, we focused on finding the sum of squares of n successive integers starting from a specific number. To simplify, we rely on a particular formula used for arithmetic sequences:

• The sum of squares formula for the first k integers is:

\[\sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}\]This formula gave us the sum of squares from the beginning integer, a, to the end, a+n-1. We calculated according to the principle that squares add a non-linear dimension to the result, different from a simple arithmetic mean.
This approach illustrates why specific formulas are needed according to the type of series being analyzed.

Independence of Variables

Independence of variables refers to a scenario in which the outcome does not rely on certain initial values. In this case, when calculated correctly, the expression \(nt - s^2\) for the given sum of integers and their squares is independent of the starting integer 'a'.
Initially, each term in our final expression relating to 'a' canceled out when simplified, indicating that their impact was neutralized.
By expanding and simplifying the terms from our derived formula:

• We obtained that all terms involving 'a' were eliminated, demonstrating the independent nature of the entire expression from 'a'.

It showed that \(nt - s^2\) resulted in a constant reliant only on 'n', the number of terms—not on their starting point.
This concept highlights the power of algebraic expressions and simplifications, showing how transformations can break dependencies on initial parameters. Such independence is crucial when analyzing theoretical models and explaining general principles in mathematics.