Find-turn-it-into-non-segma-expression-1-k-1-n-1-1-k-3-2-

Question Number 74573 by Raxreedoroid last updated on 26/Nov/19

Find (turn it into non - segma expression)

1 + \underset{k = 1}{\sum^{n - 1}} \frac{{(- 1)}^{k} + 3}{2}

Commented by mathmax by abdo last updated on 26/Nov/19

l e t S = 1 + \sum_{k = 1}^{n - 1} \frac{3 + {(- 1)}^{k}}{2} = 1 + \frac{3}{2} \sum_{k = 1}^{n - 1} (1) + \frac{1}{2} \sum_{k = 1}^{n - 1} {(- 1)}^{k}

= 1 + \frac{3}{2} (n - 1) + \frac{1}{2} (\sum_{k = 0}^{n - 1} {(- 1)}^{k} - 1)

= \frac{3 n}{2} - \frac{1}{2} - \frac{1}{2} + \frac{1}{2} \times \frac{1 - {(- 1)}^{n}}{2} = \frac{3 n}{2} - 1 + \frac{1}{4} (1 - {(- 1)}^{n})

= \frac{3 n}{2} - \frac{3}{4} - \frac{{(- 1)}^{n}}{4}

Answered by mind is power last updated on 26/Nov/19

1 + \underset{k = 1}{\sum^{n - 1}} \frac{{(- 1)}^{k}}{2} + \underset{k = 1}{\sum^{n - 1}} \frac{3}{2}

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

= \frac{{(- 1)}^{n - 1} + 6 n - 3}{4}

Commented by Smail last updated on 26/Nov/19

1 - x^{n} = (1 - x) (1 + x + x^{2} + x^{3} + \dots + x^{n - 1})

\Leftrightarrow \frac{1 - x^{n}}{1 - x} = 1 + x + x^{2} + \dots + x^{n - 1}

T h u s, \underset{k = 0}{\sum^{n - 1}} x^{k} = \frac{1 - x^{n}}{1 - x}

Commented by JDamian last updated on 26/Nov/19

\underset{k = 1}{\sum^{n - 1}} x^{k} i s t h e s u m o f a g e o m e t r i c p r o g e s s i o n

Commented by Raxreedoroid last updated on 26/Nov/19

What the step of expanding \underset{k = 1}{\sum^{n - 1}} \frac{{(- 1)}^{k}}{2} called ? or how it was done ?