Question

The value of ${\sum }_{r=1}^n$(6$r^2$ - 4r +3) is

• A. n(2n + 5)
• B. $n^2$(2n + 5)
• C. n$(2n+5{)}^2$
• D. $n\left[2n^2+n+2\right]$

Answer 1

The correct option is D $n\left[2n^2+n+2\right]$

${\sum }_{r=1}^n\left(6r^2-4r+3\right)$

$\frac{6\times n\left(n+1\right)\left(2n+1\right)}{6}-4\frac{n\left(n+1\right)}{2}+3n$

$n\left(n+1\right)\left(2n+1\right)-2n\left(n+1\right)+3n$

$n\left[2n^2+3n+1-2n-2+3\right]$

$n\left[2n^2+n+2\right]$

Hence,

option $\left(D\right)$ is correct answer.