It is given that S_n =Σ_(r=1) ^n (3r^(2 ) −3r−1). Use the the formulae of Σ_(r=1) ^n r^(2 ) and Σ_(r=1) ^n r to show that S


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Question Number 63428 by Rio Michael last updated on 04/Jul/19

$I t i s g i v e n t h a t S_{n} = \underset{r = 1}{\sum^{n}} (3 r^{2} - 3 r - 1) . U s e t h e t h e f o r m u l a e$ $o f \underset{r = 1}{\sum^{n}} r^{2} a n d \underset{r = 1}{\sum^{n}} r t o s h o w t h a t S_{n} = n^{3} .$ $s i r F o r k u m M i c h a e l$
Commented by Tony Lin last updated on 04/Jul/19

$\underset{r = 1}{\sum^{n}} (3 r^{2} - 3 r - 1)$ $= 3 \underset{r = 1}{\sum^{n}} r^{2} - 3 \underset{r = 1}{\sum^{n}} r - \underset{r = 1}{\sum^{n}} 1$ $= \frac{3 n (n + 1) (2 n + 1)}{6} - \frac{3 n (n + 1)}{2} - n$ $= \frac{n (n + 1) (2 n + 1) - 3 n (n + 1) - 2 n}{2}$ $= \frac{(2 n^{3} + 3 n^{2} + n) - (3 n^{2} + 3 n) - 2 n}{2}$ $\Rightarrow t h e q u e s t i o n m a y b e \underset{r = 1}{\sum^{n}} (3 r^{2} - 3 r + 1)$ $t h e n \frac{(2 n^{3} + 3 n^{2} + n) - (3 n^{2} + 3 n) + 2 n}{2}$ $= n^{3}$
Commented by Rio Michael last updated on 04/Jul/19

$c o r r e c t$
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