Question

The series of natural numbers is divided into groups (1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10), _ _ _ _ _ _. Find the sum of the numbers in the ${n}^{th}$ group.

Answer 1

Hint: Here, first of all, find the first term and number of terms in the ${n}^{th}$ group by using the formula ${{a}_{n}}=a+\left( n-1 \right)d$. Now, find the sum of the terms of the ${n}^{th}$ group by using the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ where n is the number of terms, a is the first term and d is the common difference of A.P.

Complete step-by-step answer:
Here, we are given a series of natural numbers divided into groups as (1), (2, 3, 4), (3, 4, 5, 6, 7), (4, 5, 6, 7, 8, 9, 10), _ _ _ _ _ _. We have to find the sum of the numbers in the ${n}^{th}$ group. First of all, let us consider the sequence given in the question.
S: (1), (2, 3, 4), (3, 4, 5, 6, 7), _ _ _ _ _ (n, n + 1, …..)
Here, we can see that in each group, each successive term differs by 1 from the previous term. We know that if there is some common difference between the terms in the sequence, then that sequence is called Arithmetic Progression.
Here, we have to find the sum of the numbers in the ${n}^{th}$ group. We know the sum of the terms of the group in A.P is given by
$\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right].....\left( i \right)$
where ‘a’ is the first term, ‘d’ is the common difference and ‘n’ is the number of terms of this group. Now to find the sum of the terms of the ${n}^{th}$ group, first, we need to find the first term, common difference, and the number of terms of this group. Since the first term of the first group is 1, the first term of the second group is 2, the first term of the third group is 3, and so on. Therefore, we get the first term of the ${n}^{th}$ group as ‘n’.
$\underbrace{1,2,3.....n}_{\text{First terms of each group}}$
Now, we have to find the number of terms in the ${n}^{th}$ group. As we can see that the first group has 1 term, the second group has 3 terms and the third group has 5 terms. So, we will find the terms in the ${n}^{th}$ group as follows:
$\underbrace{1,3,5,7,9.....}_{\text{Number of terms in each group}}$
We can see that the above sequence is in A.P with common difference 2. So, to find the number of terms in the ${n}^{th}$ group, we have to find the ${n}^{th}$ term of the above A.P. We know that the ${n}^{th}$ term of A.P is given by ${{a}_{n}}=a+\left( n-1 \right)d$ where ‘a’ is the first term, ‘d’ is the common difference and ‘n’ is the number of terms. So by substituting a = 1, d = 2 and n = n, we get the number of terms in ${n}^{th}$ group as:
${{a}_{n}}=1+\left( n-1 \right)2$
$=1+2n-2$
$=2n-1$
Hence, we get the number of terms in the ${n}^{th}$ group as (2n – 1). Also, we know the common difference between the terms of each group is 1. So, by substituting a = n, d = 1, n = (2n – 1) in equation (i), we get, the sum of the terms of the ${n}^{th}$ group,
$S=\dfrac{\left( 2n-1 \right)}{2}\left[ 2n+\left( 2n-1-1 \right)1 \right]$
$=\dfrac{\left( 2n-1 \right)}{2}\left[ 2n+2n-2 \right]$
$=\dfrac{\left( 2n-1 \right)}{2}\left( 4n-2 \right)$
$=\left( 2n-1 \right)\left( 2n-1 \right)$
$={{\left( 2n-1 \right)}^{2}}$
$=4{{n}^{2}}+1-4n$
Hence, we get the sum of the terms of the ${n}^{th}$ group as $4{{n}^{2}}+1-4n$.

Note: In this question, some students make mistakes while substituting the value of n that is the number of terms. So this must be taken care of. Also, some students take the number of terms in the ${n}^{th}$ group as (2n + 1) instead of (2n- 1) which is wrong because here n is starting from 1 and not 0 because we don’t have 0 numbers in any group. So, these mistakes must be avoided and points should be there in mind.