Question

$1+4+10+19+...\frac{3n^2-3n+2}{2}=$

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

Answer 1

Answer: (B)

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

${\sum }_{k=1}^n\frac{3k^2-3k+2}{2}=\frac{3}{2}{\sum }_{k=1}^n{k}^2-\frac{3}{2}{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1$

$=\frac{3}{2}\frac{n(n+1)(2n+1)}{6}-\frac{3}{2}\frac{n(n+1)}{2}+n$

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

$=\frac{n(n^2+1)}{2}$