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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The sum of the first and third term of an arithmetic progression is $12$ and the product of the first and second term is $24$. Find the first term.

The $9^{th}$ term of the series $27 + 9 + 5\frac{2}{5} + 3\frac{6}{7} + \dots$ will be

The sixth term of an $A.P.$ is equal to $2$. The value of the common difference $x$ of the $A.P.$ which makes the product $a_1 a_4 a_5$ least is given by:

The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

The fifth term of the $H.P.$,$2, 2\frac{1}{2}, 3\frac{1}{3}, \dots$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module