any-hint-about-how-to-prove-by-induction-in-the-Sigma-notion-topic-like-in-

Question Number 64871 by Rio Michael last updated on 22/Jul/19

a n y h i n t a b o u t h o w t o p r o v e b y i n d u c t i o n i n t h e S i g m a n o t i o n t o p i c ?

l i k e i n Σ

Commented by mathmax by abdo last updated on 22/Jul/19

y o u q u e s t i o n i s n o t c l e a r \dots .

Commented by Rio Michael last updated on 22/Jul/19

h e l p m e k n o w h o w t o p r o v e b y m a t h e m a t i c a l i n d u c t i o n w h e n d e a l i n g w i t h s u m m a t i o n

Σ

Commented by mathmax by abdo last updated on 22/Jul/19

i d o n t u n d e r s t a n d w h a t y o u w a n t b u t i g i v e s o m e p r o p e r t i e s o f

o f Σ i f w e h a v e (x_{i})_{1 ⩽ i ⩽ n} a n d {(y_{i})}_{i \in [0, n]} 2 s e q u e n c e s o f r e a l s

n u m b e r s (o r c o m p l e x) t h e s u m x_{1} + x_{2} + \dots . + x_{k} i s w r i t e n

\sum_{i = 1}^{k} x_{i} a l s o w e h a v e \sum_{i = 1}^{n} x_{i} \bar{+} \sum_{i = 1}^{n} y_{i} = \sum_{i = 1}^{n} (x_{i} \bar{+} y_{i})

\sum_{i = 1}^{n} λ x_{i} = λ \sum_{i = 1}^{n} x_{i} (λ \in R o r C)

\sum_{i = 1}^{n} (1) = n, \sum_{i = 0}^{n} (1) = n + 1, \sum_{i = 1}^{n} λ = n λ

\sum_{i = 0}^{n} λ = (n + 1) λ, \sum_{i} \sum_{j} x_{i j} = \sum_{j} \sum_{i} x_{i j} \dots .

Commented by Rio Michael last updated on 22/Jul/19

w e l l t h a n k s f o r t h a t

Commented by mathmax by abdo last updated on 23/Jul/19

y o u a r e w e l c o m e .