Question

http://prntscr.com/kn1c6d

Answer 1

Expand the brackets, lift any constants outside the summations, and use these:

\( \sum _{r=1}^{n}1=n \)


\( \sum _{r=1}^{n}r={\frac {1}{2}}n(n+1) \)


\( \sum _{r=1}^{n}r^{2}={\frac {1}{6}}n\left(n+1\right)\left(2n+1\right) \)

Answer 2

Owh u mean expand the brackets for
\((2r-1)^2\)
If so,I will get \(4r^2-4r+1\)

Answer 3

then,I will need to substitute \(\color{red}{r}\) and \(\color{red}{r^2}\) into
\(4\color{red}{r^2}-4\color{red}{r}+1\)

Answer 4

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

Answer 5

Is it correct? :O
Sorry,Im new to sum of series
Trying to finish maths b4 I go to uni

Answer 6

Typo

Plus apply the first rule too: \(\sum _{r=1}^{n}1=n\)

All marked in red...

\(\color{red}{\frac{2}{3}}n(n+1)(2n+1)-2n(n+1)+\color{red}{n}\)

I get 1330

Answer 7

Oooo,I c
yes,Im halfway right!
Yeah,I got 1330 too
Thank you @sillybilly123 ^^^