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Question: Find the sum of the products of every pair of the first \(n\) natural numbers. \(\left( a \right){...

Find the sum of the products of every pair of the first nn natural numbers.
(a) S = (n−1)2(3n+2)24\left( a \right){\text{ S = }}\dfrac{{{{\left( {n - 1} \right)}^2}\left( {3n + 2} \right)}}{{24}}
(b) S = n(n+1)(n−1)(3n−2)24\left( b \right){\text{ S = }}\dfrac{{n\left( {n + 1} \right)\left( {n - 1} \right)\left( {3n - 2} \right)}}{{24}}
(c) S = n(n−1)2(3n+2)24\left( c \right){\text{ S = }}\dfrac{{n{{\left( {n - 1} \right)}^2}\left( {3n + 2} \right)}}{{24}}
(d) S = n(n+1)(n−1)(3n+2)24\left( d \right){\text{ S = }}\dfrac{{n\left( {n + 1} \right)\left( {n - 1} \right)\left( {3n + 2} \right)}}{{24}}

Explanation

Solution

First of all we will find the sum of the first nnnatural number and then we will find the sum of the square of the first nnnatural number. And then lastly we will calculate the cube of the first nnnatural number. After that, we will combine them to find the sum of the product of every possible combination. In this, we can get the solution.

Formula used:
Sum of the first nnnatural number, S=n(n+1)/2S = n\left( {n + 1} \right)/2
The sum of the square of the first nnnatural number will be n(n+1)(2n+1)/6n\left( {n + 1} \right)\left( {2n + 1} \right)/6
The sum of the cube of the first nnnatural number will be S2{S^2}
Here,
nn, will be numbered.

Complete step by step solution:
As we have seen in the formula, so from this the sum of the product of every possible combination of the first nnnatural number is as-
⇒1.2+1.3+1.4+......+2.3+2.4+.....3.4+3.5+....+(n−1)n\Rightarrow 1.2 + 1.3 + 1.4 + ...... + 2.3 + 2.4 + .....3.4 + 3.5 + .... + \left( {n - 1} \right)n
Now on solving this, by using the formula we get
⇒[1+2+3+......(n−1)+n]2\Rightarrow {\left[ {1 + 2 + 3 + ......\left( {n - 1} \right) + n} \right]^2}
On removing the square, we get
⇒12+22+32+....+(n−1)2+n2+2(1.2+1.3+1.4+...+2.3+2.4+....(n−1)n)\Rightarrow {1^2} + {2^2} + {3^2} + .... + {\left( {n - 1} \right)^2} + {n^2} + 2\left( {1.2 + 1.3 + 1.4 + ... + 2.3 + 2.4 + ....\left( {n - 1} \right)n} \right)
On applying the summation, we get
⇒(∑n)2=∑n2+2s\Rightarrow {\left( {\sum n } \right)^2} = \sum {{n^2} + 2s}
Now from here, SSwill be calculated as
⇒S=(∑n)2−∑n2+2s2\Rightarrow S = \dfrac{{{{\left( {\sum n } \right)}^2} - \sum {{n^2} + 2s} }}{2}
Now on expanding the summation and solving it, we get
\Rightarrow \dfrac{{{{\left\\{ {n\left( {n + 1} \right)} \right\\}}^2} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{2}
Now on further solving more, we will get
⇒n2(n+1)24−n(n+1)(2n+1)62\Rightarrow \dfrac{{\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{2}
And on the calculation, we will get
⇒n(n+1)(3n2−n−2)24\Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {3{n^2} - n - 2} \right)}}{{24}}
And it can also be written as
 S = n(n+1)(n−1)(3n+2)24{\text{ S = }}\dfrac{{n\left( {n + 1} \right)\left( {n - 1} \right)\left( {3n + 2} \right)}}{{24}}

Hence,  n(n+1)(n−1)(3n+2)24{\text{ }}\dfrac{{n\left( {n + 1} \right)\left( {n - 1} \right)\left( {3n + 2} \right)}}{{24}}will be the sum of the product of every possible combination
Therefore, the option (d)\left( d \right)will be correct.

Note:
This type of question can be solved by using the simple formula which we had seen above. For this, we have to remember the formula, and also we should have the knowledge of solving the equation. Therefore, in short, I would suggest that we should always be careful while solving this type of problem as it requires very more calculations.