Sum of powers of i with odd exponents from 1 to 555

[Callmelucky]

Homework Statement

How can I solve $1+i^3+i^5+...+i^{553}+i^{555} \ ?$

Relevant Equations

i^2=-1

How can I solve 1+i^3+i^5+...+i^553+i^555? I know how to solve 1+i+i^2+i^3+i^4...+i^1001(first 4 when added =0 and then we have 1002 members, we divide that by 4, get 250 groups of 0+0+0... which is 0 and then last two members are i^1000 + i^1001 which = 1+i) and such problems but I am confused with this when the difference between exponents is 2 and I have an odd number in the last exponent.

Thank you.

 

[fresh_42]

Homework Statement:: How can I solve 1+i^3+i^5+...+i^553+i^555?
Relevant Equations:: i^2=-1

How can I solve 1+i^3+i^5+...+i^553+i^555? I know how to solve 1+i+i^2+i^3+i^4...+i^1001(first 4 when added =0 and then we have 1002 members, we divide that by 4, get 250 groups of 0+0+0... which is 0 and then last two members are i^1000 + i^1001 which = 1+i) and such problems but I am confused with this when the difference between exponents is 2 and I have an odd number in the last exponent.

Thank you.

$i^2 =-1$ so $i^4 = 1$. So $i^5= i^4 \cdot i = 1\cdot i = i.$ We need to know all odd powers: $i^{4n+1}$ and $i^{4n+3}.$

 

[pasmith]

Homework Statement:: How can I solve 1+i^3+i^5+...+i^553+i^555?
Relevant Equations:: i^2=-1

How can I solve 1+i^3+i^5+...+i^553+i^555? I know how to solve 1+i+i^2+i^3+i^4...+i^1001(first 4 when added =0 and then we have 1002 members, we divide that by 4, get 250 groups of 0+0+0... which is 0 and then last two members are i^1000 + i^1001 which = 1+i) and such problems but I am confused with this when the difference between exponents is 2 and I have an odd number in the last exponent.

You have described a geometric series (except possibly for the first term being 1 rather than $i$). The formula $$\sum_{k=0}^n ar^n = \frac{a(1 - r^{n+1})}{1 - r}$$ holds for $a \in \mathbb{C}$ and $r \in \mathbb{C} \setminus \{1\}$. Set $a = i$ and $r = i^2$ and see what happens.

 

[Steve4Physics]

$S = 1+i^3+i^5+...+i^{553}+i^{555}$

The second term is $i^3=-i$ and the $n^{th}$ term (for $n \ge 3$) is $i^2 (=-1)$ times the $(n-1)^{th}$ term. So $S$ can be written as:

$S = 1 - i + (-1)(-i) + (-1)(-1)(-i) + (-1)(-1)(-1)(-i) + …$
$= 1 - i + i - i + i …$

Now work out the number of terms...

Edit: cosmetic changes only.

 

[Callmelucky]

thank you