Problem 257 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Sum the series \(1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots \cdots \cdot\) to \(n\) terms.

Step-by-Step Solution

Answer

The sum of the series \(1+(1+2)+(1+2+3)+\cdots\) up to \(n\) terms is given by the closed-form formula: \[S_n = \frac{n(n^2 + 1)}{6}\]

Step 1: Identify the pattern of triangular numbers

Each term in the series is a sum of consecutive integers, which can be represented by the triangular numbers \(T_k\). The first few triangular numbers are \(T_1 = 1\), \(T_2 = 1 + 2 = 3\), \(T_3 = 1 + 2 + 3 = 6\), and so on. The series to sum can be represented as \(T_1 + T_2 + T_3 + \cdots + T_n\).

Step 2: Find the general formula for a triangular number

The \(k\)-th triangular number can be calculated using the formula: \[T_k = \frac{k(k + 1)}{2}\]

Step 3: Substitute the formula for triangular numbers into the series

Replace each term in the series with its triangular number formula: \[\sum_{k=1}^n T_k = \sum_{k=1}^n \frac{k(k + 1)}{2}\]

Step 4: Distribute the summation over each term

Distribute the summation over each term and simplify: \[\sum_{k=1}^n \frac{k(k + 1)}{2} = \frac{1}{2}\sum_{k=1}^n (k^2 + k)\]

Step 5: Split the summation and use summation formulas

Split the summation into two parts, apply known summation formulas for the sum of squares and sum of natural numbers, and simplify: \[\frac{1}{2}\left(\sum_{k=1}^n k^2 + \sum_{k=1}^n k\right) = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)\]

Step 6: Combine terms and simplify

Combine the terms and simplify the expression to obtain the closed-form formula for the sum of the series: \[S_n = \frac{n(n^2 + 1)}{6}\]

Key Concepts

Triangular Numbers

Triangular numbers are a series of numbers that form triangular patterns when organized as dots. To visualize this, imagine a series of dots arranged in an equilateral triangle. The first level has just one dot, the second level has two dots, and so forth, until the nth level. The total number of dots up to a certain level gives us a triangular number.

Triangular Numbers in Series

As seen in the exercise, \(T_1 = 1\), \(T_2 = 1 + 2\), \(T_3 = 1 + 2 + 3\), etc., represent the sum of consecutive integers leading up to the given term. One can easily calculate the nth triangular number using the formula \[T_n = \frac{n(n + 1)}{2}\], which neatly captures the pattern in a simple expression. In our example, the series consists of triangular numbers up to the nth term, and hence understanding this concept is crucial for solving the exercise.

Summation Formulas

Summation formulas are powerful tools in mathematics for finding the sum of sequences. These formulas provide shortcuts for calculating the sum of a series quickly without having to manually add each term.

Common Summation Formulas

For instance, to find the sum of the first n natural numbers, we use the formula \[\sum_{k=1}^n k = \frac{n(n + 1)}{2}\], while the sum of the squares of the first n natural numbers is given by \[\sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}\]. These formulas are essential for solving problems involving series, as shown in Step 5 of the solution where the summation of k and k squared are split and then calculated using the summation formulas.

Series and Sequences

In mathematics, series and sequences are related concepts that deal with ordered lists of numbers. A sequence is a list of numbers in a certain order, while a series is the sum of the terms of a sequence.

Understanding Series

A familiar example of a series is the sum of a sequence of natural numbers: 1, 2, 3, ... , n. The series can be written as \(1 + 2 + 3 + \cdots + n\), and the process of finding its sum is called summation. Whenever we encounter a series, we often look for a pattern or a formula that describes the terms in the series, which then helps us find their sum. The exercise deals with summing a particular kind of series where each term itself is a sum, forming triangular numbers.

Closed-Form Formula

A closed-form formula is an exact expression that allows us to calculate the value of a series without directly performing all of the additions. This type of formula is highly sought after in mathematics as it provides a quick and elegant solution to summing series.

Simplified Results with Closed-Formulas

The ultimate goal is to arrive at an expression where we can plug in the value of n to directly get the sum of the series up to that term. In the final step of our problem-solving, we've arrived at the closed-form formula \[S_n = \frac{n(n^2 + 1)}{6}\] for the sum of the series, demonstrating the power of closed-formulas in efficiently calculating the sum for any value of n without lengthy computations.