Question

The series of natural numbers is divided into groups $(1);{\text{ }}(2,3,4);{\text{ }}(5,6,7,8,9);.......$ and so on. Show that the sum of the numbers in nth groups is ${(n - 1)^3} + {n^3}$ .

Answer 1

Here we use the formula of the sum of an Arithmetic Progression.
And the formula is $s = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$ where $n$ is the number of terms in an A.P., $a$ is the first term of that A.P., and $d$ is the common difference.

Complete step by step answer:
Here, all terms in a group are in A.P. with common difference 1. So $d = 1$ .
When you observe, the first group contains 1 term, the second group contains 3 terms, the third group contains 5 terms and so on. So the ${n^{th}}$ group contains $(2n - 1)$ terms.
On observing all terms of all the groups, we can find that the last term of every group is a perfect square. For example, the last term of the second group is 4, which is square of 2. Similarly, the last term of the third group is 9, which is a square of 3. So the last term of ${n^{th}}$ group is ${n^2}$ . So the first term of a group is, last term of its preceding group plus 1.
So we get first term of every group as, ${(n - 1)^2} + 1$
So for ${n^{th}}$ group, number of terms is $(2n - 1)$ and first term is ${(n - 1)^2} + 1$
Now substituting these values in sum of an A.P. formula, we get,
$s = \dfrac{{(2n - 1)}}{2}\left( {2({{(n - 1)}^2} + 1) + (2n - 1 - 1)} \right)$ .(as $d = 1$ )
$\Rightarrow s = \dfrac{{(2n - 1)}}{2}\left( {2{n^2} - 4n + 4 + 2n - 2} \right)$
On taking 2 as common from the second term and canceling that with the 2 in denominator, we get,
$\Rightarrow s = (2n - 1)({n^2} - n + 1)$
$\Rightarrow s = 2{n^3} - 3{n^2} + 3n - 1$
Here, we can split the $1^{st}$ term into two halves. So,
$\Rightarrow s = {n^3} + {n^3} - 3{n^2} + 3n - 1$
When you observe, we can group terms like this.
$\Rightarrow s = {n^3} + ({n^3} - 3{n^2} + 3n - 1)$
Which results in,
$\Rightarrow s = {n^3} + {(n - 1)^3}$
and we all know the identity that ${(n - 1)^3} = {n^3} - 3{n^2} + 3n - 1$
We have shown that the sum of terms in the nth group is ${n^3} + {(n - 1)^3}$.

Note:
Here, we have implemented a technique where the first term of a group is the next term of its preceding group’s last term. But rather, we can also find the sum in another way. Suppose that we have got the last term of the nth group. Then consider it as the first term of that A.P., and take the common difference as $- 1$ , and substitute the values in the sum formula. In either way, you will get the same answer.