Problem 253 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The short answer for the sum of the series up to 'n' terms is: \( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \)

Step 1: Identify the pattern in the series

First, observe the given series and the terms, which are in the form of (k)(k+1)^2. This helps us to understand the pattern in the series, and set up the expression for the partial sum.

Step 2: Write the expression for partial sum

Now, let's write the expression for the partial sum (Sn) of the series up to n terms: \( S_n = \sum_{k=1}^{n} (k)(k+1)^2 \)

Step 3: Find the general formula for partial sum

To find the formula for Sn, we'll break down the terms in the series: \( S_n = \sum_{k=1}^{n} (k)(k^2 + 2k + 1) \) Now, distribute k to each term inside the parentheses and then split the series sum: \( S_n = \sum_{k=1}^{n}(k^3 + 2k^2 + k) \) \( S_n = \sum_{k=1}^{n}k^3 + \sum_{k=1}^{n}2k^2 + \sum_{k=1}^{n}k \) We can use the formulas for the sum of the cubes, sum of the squares, and sum of the first 'n' natural numbers. \( \sum_{k=1}^{n}k^3 = (\frac{n(n+1)}{2})^2 \) \( \sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6} \) \( \sum_{k=1}^{n}k = \frac{n(n+1)}{2} \)

Step 4: Substitute and simplify the expression

Now, substitute these expressions into the partial sum of the series: \( S_n = (\frac{n(n+1)}{2})^2 + 2(\frac{n(n+1)(2n+1)}{6}) + \frac{n(n+1)}{2} \) Simplify the expression: \( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \)

Step 5: Final answer

The expression we found in step 4 is the general formula for the sum of the given series up to 'n' terms: \( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \) This formula can be used to find the sum of the series up to any number of terms 'n'.

Key Concepts

Arithmetic Series

An arithmetic series is a sequence of numbers in which each term after the first is the sum of the previous term and a constant, known as the common difference. For example, the sequence 2, 4, 6, 8 is an arithmetic series where the common difference is 2.
In a general arithmetic series, the first term is denoted by \(a\), and the common difference is \(d\). The nth term, \(a_n\), can be found using the formula:

• \(a_n = a + (n-1)d\)

The sum of the first n terms of an arithmetic series, \(S_n\), can be calculated using:

• \(S_n = \frac{n}{2} \times (2a + (n-1)d)\)

This is helpful when dealing with series where differences between consecutive terms remain constant.

Sum of Cubes

The sum of cubes is a mathematical concept where you calculate the sum of all the cubes of the first n natural numbers. This is given by the formula:

• \(\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2\)

The formula is derived from the binomial theorem and works beautifully because it shows that the sum of cubes is actually the square of the sum of the first natural numbers. It's an elegant representation showing how numbers can relate to each other in unexpected ways.
Understanding the sum of cubes helps in solving various problems related to series summation, as demonstrated in the original exercise.

Sum of Natural Numbers

The sum of the first n natural numbers is one of the fundamental ideas in arithmetic, used frequently in many areas of mathematics. The formula to calculate this sum is:

• \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)

The derivation of this formula has a story rooted in legend, where young Carl Friedrich Gauss unexpectedly discovered it to efficiently solve a classroom sum. This formula shows a simple arithmetic progression and gives a straightforward way to calculate the total of a consecutive series, playing a crucial role in many complex equations like those involving series summations.

Sum of Squares

The sum of the squares of the first n natural numbers is a bit more complex than the sum of the numbers themselves. The formula for this is:

• \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\)

This formula helps find the total of square values from 1 through n. It's widely used in statistics, physics, and engineering, especially when dealing with variance and standard deviation where squared terms often appear. The sum of squares represents the idea of magnifying differences by squaring, which is critical in assessing data distribution and potential outliers. This concept was key in solving the original exercise, highlighting its necessity in summation problems.