1. ## Series

How would I write this series with the $\Sigma$ sign and all:

25 (1 + 2.04 + 3.04 + 4.08 + 5.08 + 6.12 + 7.12 + 8.16 ...)

OR

25 + 51 + 76 + 102 + 127 ....

2. $\sum_{n\,=\,0}^k25\left(n+0.04\left\lfloor\frac n2\right\rfloor\right)$

$=\quad\sum_{n\,=\,0}^k\left(25n+\left\lfloor\frac n2\right\rfloor\right)$

3. Right I see the abstractionist is also looking at this thread so by the time i post this he''ll have a better answer down probably...

Do it as two sums...

$\bigg{(} \bigg{(} 25 \sum_{n=0}^\infty 2n \bigg{)} + n \bigg{)} + \bigg{(} \bigg{(} 25 \sum_{n=0}^\infty (2n+1) \bigg{)} + n \bigg{)}$

4. Hello, metlx!

Write in sigma-notation:

$(a)\;\;S \;=\;25\,\bigg(1 + 2.04 + 3.04 + 4.08 + 5.08 + 6.12 + 7.12 + 8.16 + \hdots\bigg)$
This is quicky tricky . . .

$\text{We have: }\;S \;=\;\;25\bigg(1 + \underbrace{2.04 + 3.04} + \underbrace{4.08 + 5.08} + \underbrace{6.12 + 7.12} +$ $\underbrace{8.16 + 9.16} + \hdots\bigg)$

. . . . . . $S \;=\;25\bigg(1 + 5.08 + 9.16 + 13.24 + 17.32 + \hdots\bigg)$

. . . . . . $S \;=\;25\bigg([1 + 0(0.8)] + [5 + 1(0.08)] + [9 + 2(0.8)] + [13 + 3(0.8)] + \hdots\bigg)$

. . . . . . $S \;=\;25\sum_{k=0}^n\bigg((4k+1) + k(0.8)\bigg) \;=\;25\sum_{k=0}^n\bigg(4.8k + 1\bigg)$

$(b)\;\;S \;=\;25 + 51 + 76 + 102 + 127 + \hdots$
$\text{We have: }\;S \;=\;\underbrace{25 + 51} + \underbrace{76 + 102} + \underbrace{127 + 153} + \underbrace{178 + 204} + \hdots$
. . . . . . $S \;=\;76 + 178 + 280 + 382 + \hdots$ . which has a common difference of 102
Therefore: . $S \;=\;\sum_{k=0}^n\bigg(76 + 102k\bigg)$