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Assuming that 'a' is real, a^5 = 1 implies that a = 1. but it had been shown that 'a' not equal 1. by inspection, the solution for real 'a' is approximately slightly less than -1 (somewhere between -2 and -1).
Consider a geometic series with 1st element 1 and ratio a. The nth element 0:06 is a^(n-1). Given a⁴+a³+a²+a+1=0 Multiply by a: a+a²+a³+a⁴+a⁵=0. It is clear sum of any 5 consecutive element is 0. a⁵[1+(1/a)+(1/a²)+(1/a³)+(1/a⁴)=0 Consider k=a²⁰¹⁰+a²⁰⁰⁹+...+a⁵+a⁴+a³+a²+a+1 There are 2011 elements. Counting from left 5 elements each time, k=1. But counting each time 5 elements from right, k=a²⁰¹⁰ Thus a²⁰¹⁰=1 In the same way, a²⁰²⁰=1. Therefore a²⁰²⁰+a²⁰¹⁰+1=3
a = 1 doesn't satisfy the equation hence (a -1) is not equal to zero and we can multiply both sides by (a-1) Multiplying both sides by (a -1) we get a^5 = 1 putting in expression Result = 1+1+1 = 3
1+a+a^2+a^3+a^4=0 | recognize the general form (geometric series): sum_(k=0)^(n) a^k=1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1) so multiply both sides: (a-1)(1+a+a^2+a^3+a^4)=0 thus: a^5-1=0 So a^5=1 Thus a^2020+a^2010+1= (a^5)^(404)+(a^5)^(402)+1= 1^404+1^402+1= 1+1+1= 3
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Assuming that 'a' is real, a^5 = 1 implies that a = 1. but it had been shown that 'a' not equal 1. by inspection, the solution for real 'a' is approximately slightly less than -1 (somewhere between -2 and -1).
Consider a geometic series with 1st element 1 and ratio a. The nth element 0:06 is a^(n-1).
Given a⁴+a³+a²+a+1=0
Multiply by a: a+a²+a³+a⁴+a⁵=0.
It is clear sum of any 5 consecutive element is 0.
a⁵[1+(1/a)+(1/a²)+(1/a³)+(1/a⁴)=0
Consider
k=a²⁰¹⁰+a²⁰⁰⁹+...+a⁵+a⁴+a³+a²+a+1
There are 2011 elements. Counting from left 5 elements each time, k=1. But counting each time 5 elements from right, k=a²⁰¹⁰
Thus a²⁰¹⁰=1
In the same way, a²⁰²⁰=1.
Therefore a²⁰²⁰+a²⁰¹⁰+1=3
a = 1 doesn't satisfy the equation hence (a -1) is not equal to zero and we can multiply both sides by (a-1)
Multiplying both sides by (a -1) we get a^5 = 1
putting in expression
Result = 1+1+1 = 3
I have another solution
Action 1:
a^4 + a^3 + a^2 + a + 1 = 0
a^2 * (a+1) + a + 1 = -(a^4)
(a^2 + 1) * (a+1) = -(a^4).
Action 2:
a^4 + a^3 + a^2 + a + 1 = 0
a^2 * (a^2 + 1) + a * (a^2 + 1) = -1
a * (a+1) * (a^2 + 1) = -1
(a+1) * (a^2 + 1) = -(1/a).
And clearly 'a' can't equal zero. So we put Action 1 and Action 2 together...
Action 3:
a^4 = 1/a
a^5 = 1.
and so on, then it's the same way you did.
(a^5)^402 + (a^5)^404 + 1 = 3.
1+a+a^2+a^3+a^4=0 | recognize the general form (geometric series): sum_(k=0)^(n) a^k=1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1) so multiply both sides:
(a-1)(1+a+a^2+a^3+a^4)=0
thus:
a^5-1=0
So a^5=1
Thus a^2020+a^2010+1=
(a^5)^(404)+(a^5)^(402)+1=
1^404+1^402+1=
1+1+1=
3
2a^20 1a^10 a^5^5a^2^3^2^3 a^2^1^1^3 a^2^3 (a ➖ 3a+2). 2a^8060 2a^8^6 2a^2^3^2^3 1a^1^1^1^2^3a^2^3(a ➖ 3a+2).
is it 3?
just watch the video
Yes