The sum of all the products of the first n na | vedclass

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Question

(A) We know that the square of the sum of the first $n$ natural numbers is given by:${\left\{ \frac{n(n + 1)}{2} \right\}^2} = {(1 + 2 + \dots + n)^2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that the square of the sum of the first $n$ natural numbers is given by:${\left\{ \frac{n(n + 1)}{2} \right\}^2} = {(1 + 2 + \dots + n)^2} = \sum_{r=1}^n {r^2} + 2\sum_{1 \le s Therefore,the sum of products taken two at a time is:$\sum_{1 \le s Substituting the formula for the sum of squares $\sum_{r=1}^n {r^2} = \frac{n(n + 1)(2n + 1)}{6}$:$= \frac{1}{2} \left[ \frac{n^2(n + 1)^2}{4} - \frac{n(n + 1)(2n + 1)}{6} \right]$$= \frac{n(n + 1)}{2} \left[ \frac{n(n + 1)}{4} - \frac{2n + 1}{6} \right]$$= \frac{n(n + 1)}{2} \left[ \frac{3n^2 + 3n - 4n - 2}{12} \right]$$= \frac{n(n + 1)(3n^2 - n - 2)}{24}$$= \frac{n(n + 1)(n - 1)(3n + 2)}{24}$

The sum of all the products of the first $n$ natural numbers taken two at a time is

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