n-1-10-i-n-i-n-1-

Question Number 108825 by ajfour last updated on 19/Aug/20

\underset{n = 1}{\sum^{10}} (i^{n} + i^{n + 1}) = ?

Commented by ajfour last updated on 19/Aug/20

t h a n k s e v e r y o n e w h o c o n f i r m e d .

Answered by john santu last updated on 19/Aug/20

\frac{⊸ J S ⊸}{♡}

\underset{n = 1}{\sum^{10}} i^{n} (1 + i) = (1 + i) \underset{n = 1}{\sum^{10}} i^{n} = (1 + i) (i - 1) = - 1 - 1 = - 2

\underset{k = 1}{\sum^{10}} i^{n} = i - 1 - i + 1 + i - 1 - i + 1 + i - 1

= i - 1 (*)

Commented by ajfour last updated on 19/Aug/20

\underset{k = 1}{\sum^{10}} i^{n} = i - 1 - i + 1 + i - 1 - i + 1 + i - 1

= i - 1

s o {\sum_{n = 1}}^{10} i^{n} (1 + i) = (i + 1) (i - 1)

= - 1 - 1 = - 2 .

{i s n}^{'} t t h i s r i g h t ?

Commented by bemath last updated on 19/Aug/20

y e s

Commented by john santu last updated on 19/Aug/20

y e s

Answered by Dwaipayan Shikari last updated on 19/Aug/20

i \frac{i^{10} - 1}{i - 1} + i^{2} \frac{i^{10} - 1}{i - 1} = i (i + 1) - (i + 1) = - 1 - 1 = - 2