Question

Sum of products of first $n$ natural numbers taken two at a time is

• A. $\frac{\mathrm{n}({\mathrm{n}}^2-1)(3\mathrm{n}+2)}{24}$
• B. $\frac{\mathrm{n}{(\mathrm{n}+1)}^2(3\mathrm{n}+2)}{74}$
• C. $\frac{{\mathrm{n}}^2(\mathrm{n}+1)(3\mathrm{n}+2)}{48}$
• D. $\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(3\mathrm{n}+2)}{96}$

Answer 1

The correct option is A

$\frac{\mathrm{n}({\mathrm{n}}^2-1)(3\mathrm{n}+2)}{24}$

Explanation for the correct option:

Finding the sum of $n$ natural numbers taken two at a time:

We know that,

$$\begin{gathered} \left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+.......+{\mathrm{x}}_{\mathrm{n}}\right)={{\mathrm{x}}_1}^2+{{\mathrm{x}}_2}^2+............+{{\mathrm{x}}_{\mathrm{n}}}^2+2\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+.........+{\mathrm{x}}_{\mathrm{n}-1}{\mathrm{x}}_{\mathrm{n}}\right) \\ \mathrm{On}\mathrm{substituting}{\mathrm{x}}_1=1,{\mathrm{x}}_2=2,{\mathrm{x}}_3=3,........,{\mathrm{x}}_{\mathrm{n}}=\mathrm{n},\mathrm{we}\mathrm{get} \\ {\left(1+2+3+.......+\mathrm{n}\right)}^2=1^2+2^2+3^2+......{\mathrm{n}}^2+2\left(1.2+2.3+.........\right)\mathrm{or}{\left(1+2+3+.......+\mathrm{n}\right)}^2-\left(1^2+2^2+......{\mathrm{n}}^2\right)=2\left(\mathrm{S}\right) \\ \mathrm{When}\mathrm{S}=\mathrm{Sum}\mathrm{of}\mathrm{product}\mathrm{of}\mathrm{numbers}\mathrm{taken}\mathrm{two}\mathrm{at}\mathrm{a}\mathrm{time} \\ \mathrm{Therefore}; \\ 2\mathrm{S}={\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]}^2-\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right] \\ =\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}-\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6} \\ =\mathrm{n}\left(\mathrm{n}+1\right)\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{4}-\frac{\left(2\mathrm{n}+1\right)}{6}\right] \\ =\mathrm{n}\left(\mathrm{n}+1\right)\times \frac{1}{2}\left[\frac{3\mathrm{n}\left(\mathrm{n}+1\right)-2\left(2\mathrm{n}+1\right)}{6}\right]\left[\mathrm{Taking}\mathrm{LCM}\mathrm{and}\mathrm{taking}\frac{1}{2}\mathrm{common}\right] \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left[\frac{3{\mathrm{n}}^2+3\mathrm{n}-4\mathrm{n}-2}{6}\right] \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{12}\left(3{\mathrm{n}}^2-\mathrm{n}-2\right) \end{gathered}$$

Now finding the value of $S$

$$\begin{gathered} \mathrm{S}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{24}\left(3{\mathrm{n}}^2-3\mathrm{n}+2\mathrm{n}-2\right) \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{24}\left[3\mathrm{n}\left(\mathrm{n}-1\right)+2\left(\mathrm{n}-1\right)\right] \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}-1\right)\left(3\mathrm{n}+2\right)}{24} \\ =\frac{\mathrm{n}\left({\mathrm{n}}^2-1\right)\left(3\mathrm{n}+2\right)}{24} \end{gathered}$$

Therefore, the correct answer is option (A).