Question

The sum of the product of the integers 1,2,3,...., n taken two at a time is

• A. $\frac{n(n+1)}{24}(3n^2-n-2)$
• B. $\frac{n^2(n+1{)}^2}{2}(n^2+1)$
• C. $\frac{n(n+1)}{18}(n^2-n-2)$
• D. $\frac{n(n+1)(n+2)(n+3)}{24}$

Answer 1

The correct option is A

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

$(a+b{)}^2=(a^2+b^2)+2(ab)=\sum a^2+2\sum ab(a+b+c{)}^2=(a^2+b^2+c^2)+2(ab+bc+cd)=\sum a^2+2\sum ab\text{In general,}(a_1+a_2+\cdots +a_n{)}^2=(a_1^2+a_2^2+\cdots +a_n^2)+2(a_1{a}_2+a_1{a}_3+\cdots +a_{n-1}{a}_n)=\sum a_1^2+2\sum a_1{a}_2\therefore (1+2+3+\cdots +n{)}^2=(1^2+2^2+3^2+\cdots +n^2)+2\left(\text{sum of products of n integers taken two at a time}\right)$
i.e. $(\sum n{)}^2=\sum n^2+2S;$ where S is the required sum
$\therefore S=\frac{1}{2}\left[\right(\sum n{)}^2-(\sum n^2)]=\frac{1}{2}\left[\right\{\frac{n(n+1)}{2}{\}}^2-\left\{\frac{n(n+1)(2n+1)}{6}\right\}]=\frac{n(n+1)}{24}(3n^2-n-2)$