Question

If we wrote ${n^3}$ as the sum of three consecutive odd numbers then what will be the first term.
(A) $2n + 1$
(B) $(2n + 1)(2n - 1)$
(C) $\dfrac{{{n^3} - 6}}{3}$
(D) $n(n - 1) + 1$

Answer 1

Hint : First write one odd term. Then write a term before and after it to get three consecutive odd terms. Then add them to equate the expression with ${n^3}$ . Then solve the expression to find the value of the first odd term.

Complete step-by-step answer :
Let $a$ be some integer. Then $2a$ will be an even integer. Hence, $2a - 1$ will be an odd integer.
Thus, $2a - 3$ will be an odd integer before $2a - 1$ . And $2a + 1$ will be an odd integer after $2a - 1$ .
Therefore, $2a - 3,$ $2a - 1$ and $2a + 1$ are three consecutive odd integers.
Now, it is given in the question that, ${n^3}$ is the sum of three consecutive odd numbers.
Therefore, we can write
${n^3} = (2a - 3) + (2a - 1) + (2a + 1)$
$\Rightarrow {n^3} = 6a - 3$
Rearranging it we can write
$6a = {n^3} + 3$
Dividing both the sides by 6 we get
$a = \dfrac{{{n^3} + 3}}{6}$
We have to find the first term in term of $n$
The first term of the sequence is $2a - 3$ .
Therefore, we can write
$2a - 3 = 2\left( {\dfrac{{{n^3} + 3}}{6}} \right) - 3$
By cancelling the common terms, we get
$= \left( {\dfrac{{{n^3} + 3}}{3}} \right) - 3$
By cross multiplying, we get
$= \dfrac{{{n^3} + 3 - 9}}{3}$
$\Rightarrow 2a - 3 = \dfrac{{{n^3} - 6}}{3}$
Therefore, the first term will be $\dfrac{{{n^3} - 6}}{3}$
Therefore, from the above explanation, the correct answer is, option (C) $\dfrac{{{n^3} - 6}}{3}$

Note : Here understanding of the question was very important. You need to know how an odd number would look like and you need to know how we can write three consecutive odd terms in a general form. After that, it was a basic calculation. Don’t make the mistake of thinking that the value of $a$ is the answer. You are not asked to find the value of $a$ . You are asked to find the value of the first term which is $2a - 3$ .