Problem 78 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

There are \(n\) A.P.'s whose common difference are \(1,2,3, \ldots n\) respectively, the first term of each being unity. Prove that sum of their \(n\) th terms is \(\frac{1}{2} n\left(n^{2}+1\right)\).

Step-by-Step Solution

Answer

The sum of the nth terms of the given A.P.'s is \(\frac{1}{2}n(n^2 + 1)\).

Step 1: Find the general formula for the nth term of an A.P.

The general formula for the nth term of an arithmetic progression is given by: \(a_n = a_1 + (n-1)d\) where \(a_n\) is the nth term, \(a_1\) is the first term, n is the number of terms, and d is the common difference.

Step 2: Find the nth term of each A.P. using the general formula

For each A.P., we have to find the nth term using the general formula given above. Since there are n A.P.'s with common differences from 1 to n, we will find the nth term of each A.P. and represent them as \(a_{n1}, a_{n2}, a_{n3}, \ldots, a_{nn}\). A.P. with common difference 1: \(a_{n1} = 1 + (n-1)(1) = n\) A.P. with common difference 2: \(a_{n2} = 1 + (n-1)(2) = 2n - 1\) A.P. with common difference 3: \(a_{n3} = 1 + (n-1)(3) = 3n - 2\) Likewise, for A.P. with common difference n: \(a_{nn} = 1 + (n-1)(n) = n^2 + 1 - n\)

Step 3: Sum the nth terms of the A.P.'s

Now that we have the nth term of each A.P., we will sum them up and represent the sum as \(S_n\). \(S_n = a_{n1} + a_{n2} + a_{n3} + \ldots + a_{nn}\)

Step 4: Simplify the expression to get the desired result

We will now try to simplify the expression for \(S_n\) and get the desired result \(\frac{1}{2} n(n^2 + 1)\). \(S_n = n + (2n - 1) + (3n - 2) + \ldots + (n^2 + 1 - n)\) Notice that each term can be written as \(kn - (k-1)\), where k is the common difference. We can rewrite \(S_n\) as: \(S_n = \sum_{k=1}^{n} (kn - (k - 1))\) Now, split the summation and simplify: \(S_n = \sum_{k=1}^{n} kn - \sum_{k=1}^{n}(k - 1)\) \(S_n = n\sum_{k=1}^{n} k - \sum_{k=1}^{n}k + n\) To find the values of \(\sum_{k=1}^{n} k\) and \(\sum_{k=1}^{n}k\), we can use the formula of the sum of the first n natural numbers, which is: \(\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\) Substitute this formula in the expression for \(S_n\): \(S_n = n \cdot \frac{n(n + 1)}{2} - \frac{n(n + 1)}{2} + n\) Factor out \(\frac{n(n+1)}{2}\) which gives: \(S_n = \frac{n(n + 1)(n - 1)}{2} + n\) Now, we can factor an n out of each term: \(S_n = \frac{n^2(n + 1)(n - 1) + 2n^2}{2}\) Now, combine the terms and simplify: \(S_n = \frac{n^2(n^2 - 1) + 2n^2}{2}\) \(S_n = \frac{1}{2} n(n^2 + 1)\) Hence, the sum of the nth terms of the given A.P.'s is proved to be \(\frac{1}{2}n(n^2 + 1)\).

Key Concepts

nth term of an A.P.

Understanding the 'nth term of an A.P.' (Arithmetic Progression) is crucial for sequences and series problems in mathematics. An A.P. is a sequence of numbers in which each term after the first is found by adding a constant, known as the common difference, to the previous term.

The general formula to find the nth term () of an A.P. is given by . In this formula, is the specific term you want to find, is the first term, and d is the common difference between the terms. This formula helps us understand how the terms in an A.P. are related and how to predict any term in the sequence without listing all the terms.

summation of series

The 'summation of series' is a critical concept in understanding arithmetic progressions and how they accumulate. Summation is denoted by the Greek letter Sigma () and refers to the total value obtained by adding the terms of a series. For arithmetic progressions, the summation of the first n terms can be calculated using the formula , where n is the number of terms, is the first term, and is the nth or last term.

When dealing with series involving an arithmetic progression with varying common differences, as in our exercise example, we sum up the nth terms of each progression individually and then combine them for the grand total. This technique illustrates how each A.P. contributes to the overall sum, which exemplifies the underlying structure and relationship between individual progressions within a complex series.

mathematical induction

The principle of 'mathematical induction' is a method of mathematical proof typically used to establish that a given statement is true for all natural numbers. It is a two-step process where the first step, known as the base case, proves that the statement holds for the initial number, usually 1. The second step, called the inductive step, assumes that the statement holds for some arbitrary natural number n and then proves that the statement must also hold for n+1.

In the context of our exercise, we could use mathematical induction to prove that the developed formula for the summation of the nth terms actually holds true for any natural number n. We've illustrated a direct approach to solving our initial problem, but induction would be another valid and robust means of proof for general statements or formulas involving series and sequences.