Question

Find the sum of all the products of the first n natural numbers taken two at a time.

Answer 1

Like ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ . Similarly, ${\left( {{x_1} + {x_2} + .......... + {x_n}} \right)^2} = {x_1}^2 + {x_2}^2 + ....... + {x_n}^2 + 2{x_1}{x_2} + ............ + 2{x_{n - 1}}{x_n}$ and this equation will be used to solve the question . On further solving the values of terms $\sum n$ and $\sum {{n^2}}$ will be used so you have to remember the values to these terms. After this it will be easy to solve it further or to get the desired result.

Complete step by step answer:
In general, ${\left( {{x_1} + {x_2} + .......... + {x_n}} \right)^2} = {x_1}^2 + {x_2}^2 + ....... + {x_n}^2 + 2{x_1}{x_2} + ............ + 2{x_{n - 1}}{x_n}$
By putting ${x_1}$ $=$ $1$ , ${x_2}$ $=$ $2$ and ${x_n}$ $=$ $n$ in the above equation we get
${\left( {1 + 2 + 3 + ............ + n} \right)^2}{\text{ = }}\left( {{1^2} + {2^2} + {3^2} + ............ + {n^2}} \right) + 2\left( {{\text{sum of product of numbers taken two at a time}}} \right)$
We can also write the above equation as
${\left( {\sum n } \right)^2}{\text{ }} = {\text{ }}\sum {{n^2} + {\text{ }}2S}$
Where $S$ denotes the sum. As we know that the value of $\sum n$ is $\dfrac{{n\left( {n + 1} \right)}}{2}$ and the value of $\sum {{n^2}}$ is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ . Therefore, by substituting these values in the equation we get
$\Rightarrow {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}{\text{ }} = {\text{ }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2S$
By taking right hand side term to left hand side we get
$\Rightarrow {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}{\text{ }} - {\text{ }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ = }}2S$
Now take $\dfrac{{n\left( {n + 1} \right)}}{2}$ term common at the left hand side,
$\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }} - {\text{ }}\dfrac{{\left( {2n + 1} \right)}}{3}} \right]{\text{ = }}2S$
On solving the term inside the square brackets we get
$\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3{n^2} + 3n - 4n - 2}}{6}} \right]{\text{ = }}2S$
Taking $6$ out we get
$\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{{12}}\left[ {3{n^2} - n - 2} \right]{\text{ = }}2S$
Now shift $2$ from the right hand side to the left hand side,
$\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{{24}}\left[ {3{n^2} - n - 2} \right]{\text{ = }}S$
We can also write it as
$\Rightarrow S{\text{ = }}\dfrac{{n\left( {n + 1} \right)}}{{24}}\left[ {3{n^2} - 3n + 2n - 2} \right]{\text{ }}$
$\Rightarrow S{\text{ = }}\dfrac{{n\left( {n + 1} \right)}}{{24}}\left[ {3n\left( {n - 1} \right) + 2\left( {n - 1} \right)} \right]{\text{ }}$
Writing the equation in its factor’s form
$\Rightarrow S{\text{ = }}\dfrac{{n\left( {n + 1} \right)}}{{24}}\left[ {\left( {3n + 2} \right)\left( {n - 1} \right)} \right]{\text{ }}$
$\therefore S{\text{ = }}\dfrac{{n\left( {n + 1} \right)\left( {3n + 2} \right)\left( {n - 1} \right)}}{{24}}$

Note:
Don’t confuse $\sum n$ and $\sum {{n^2}}$ . The possible mistake that could happen is that you might think that the sum of the product of n natural numbers taken two at a time is the sum of the square of the first n natural numbers. Keep in mind the values of both these terms as they are important. It’s easy to solve the question, you just have to take a look at the starting of the solution and try to understand the concept behind this.