Question

The sum of the product of the $ 2n $ numbers $ \pm 1 $ , $ \pm 2 $ , $ \pm 3 $ ,…., $ \pm n $ taking two at a time is
(a) $ - \dfrac{{n\left( {n + 1} \right)}}{2} $
(b) $ \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $
(c) $ - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $
(d) None of these

Answer 1

Hint: To solve this problem, we will represent the number in the series form. Since the relation between the terms is already given, we can use this relation to make a series from the given numbers. Then we find the general term in all the terms of the series. After getting the general term, we can easily find the sum of the terms in the series by using predefined formulas.

Complete step-by-step answer:
It is given in the question that there are $ 2n $ numbers in the series.
If we will take given numbers $ \pm 1 $ , $ \pm 2 $ , $ \pm 3 $ ,…., $ \pm n $ twice , the series will look like
The first term will be $ {t_1} = \left( { - 1} \right) \times \left( { + 1} \right) = - 1 $ .
The second term will be $ {t_2} = \left( { - 2} \right) \times \left( { + 2} \right) = - {2^2} $ .
The third term will be $ {t_3} = \left( { - 3} \right) \times \left( { + 3} \right) = - {3^2} $ .
In a similar way we can write the $ n $ number of terms. Similarly, the last term will be, $ {t_1} = \left( { - n} \right) \times \left( { + n} \right) = - {n^2} $
We can express the series as:
 $ - 1 $ , $ - {2^2} $ , $ - {3^2} $ ,…………., $ - {n^2} $
 We can also express the terms in the above series by a general term.
 $ {t_n} = - {n^2} $
The sum of the digits can be expressed as:
 $ \sum {{t_n}} = - \sum {{n^2}} $
We know that sum of $ {n^2} $ is given as $ \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $ . Hence substituting this value in above the expression we get,
 $ \sum {{t_n}} = - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $
So, the correct answer is “Option C”.

Note: To find the sum of the series, there are certain formulas that are predefined. We can use these formulas to find the sum. In our answer the general term was $ {n^2} $ , but if the general term were $ n $ or $ {n^3} $ , we could have expressed the sum as $ \dfrac{{n\left( {n + 1} \right)}}{2} $ for $ n $ and $ \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} $ for $ {n^3} $ respectively. These predefined formulas make the calculation easy.