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( \sum_{r=1}^{n} r \right)}^2 = \left( \frac{n(n+1)}{2} \ri

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( \sum_{r=1}^{n} r \right)}^2 = \left( \frac{n(n+1)}{2} \right)^2 = \sum_{r=1}^{n} r^2 + 2 \sum_{1 \le s Rearranging the terms to find the sum of products taken two at a time:$2 \sum_{s Substituting the formula for the sum of squares $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$:$2 \sum_{s Taking $\frac{n(n+1)}{12}$ as a common factor:$2 \sum_{s $2 \sum_{s Factoring the quadratic expression $(3n^2 - n - 2) = (n-1)(3n+2)$:$2 \sum_{s Dividing by $2$:$\sum_{s < t} st = \frac{1}{24}n(n - 1)(n + 1)(3n + 2)$.

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

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