How Can I Solve ##1+i^3+i^5+...+i^{553}+i^{555}\ ?##

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To solve the series 1+i^3+i^5+...+i^553+i^555, the key is recognizing the pattern in the powers of i, which cycle every four terms. The series can be expressed as a geometric series with the first term as 1 and the common ratio as i^2. The odd powers of i alternate between -1 and i, leading to a simplification of the series. By calculating the number of terms and applying the geometric series formula, the sum can be determined. The final result will depend on the total number of terms in the series.
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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.
 
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Callmelucky said:
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}.##
 
Callmelucky said:
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 <br /> \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.
 
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##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.
 
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thank you
 
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