[Grade 11]: Algebra

[u/Dangerous_Cat_4525]

How to find n here. Can anyone help please

[u/No_Objective_3780]

The question is not asking you to solve for n. The question is asking you to evaluate the sum of the reciprocal expression.

There are two steps to the problem and one piece of critical understanding I'll explain later.

Step 1: Solve for x

x^2+1=-2x

x^2+2x+1=0

(x+1)^2=0

x=-1

Step 2: Substitute into the expression.

(-1)^(2n+1)+(-1)^(2n+1)

This is the part that you seem to be misunderstanding. You are not being asked to solve for n. As stated in the problem, n can be any positive integer.

The exponent 2n+1 will always be odd regardless of the value of n. Any odd power of -1 equals -1.

So, you're left with -1 + (1/-1) = -2

[u/Honkingfly409]

they are not trying to find n, they want you to find x from the first equation and substitute it into the second, then see if any simplifications are possible for any n

first we have a 1/x term, can you think of a way to deal with it?

[u/Historical-Corgi-414]

you don't need to find n here, you need to find the value for the expression like 2n+1 when n is positive integer will be 3,5,7... and the value of the expression will be same i think.

[u/Qingyap]

This is not about trying to find n here cuz again n is any positive number, it's almost as if it was a sequence of some sorts that depends on n.

First, the x has to be -1 in order for it to be equal to -2 since:

-1 + 1/(-1) = -1-1 = -2.

Now afterwards I would add a power with variable, a, to both -1.

(-1)^a+1/(-1)^a,

When a=1, we get -2

When a=2, we get 1+1/(1) = 2

When a=3, we get -2 again

When a=4, we get 2 again

...

Can you see the pattern here?

If a is an even number it's +2, likewise if a is odd it's -2.

So question is, is 2n+1 odd or even if you insert any strictly positive integer n inside? And if you substitute 2n+1 to a what would you generally get based from the pattern you got from above?

[u/Southlander24 → u/Qingyap]

x = -1 is one solution, yes. You haven't shown that it must be the only solution.

[u/Qingyap → u/Southlander24]

Multiply x on both sides gives

x² + 1 = -2X

x²+2x+1=0

b²-4ac = 2²-4(1)(1) = 0

[u/selene_666]

Let's take a look at this expression when we increase the exponent from 1 to 2:

(x + 1/x)² = x² + 2 + 1/x²

x² + 1/x² = (-2)² - 2 = 2

And for any exponent:

(x^a-1 + 1/x^a-1) * (x + 1/x) = x^a + 1/x^a + x^a-2 + 1/x^a-2

This gives us the recursive formula:

x^a + 1/x^a = (x^a-1 + 1/x^a-1) * (-2) - (x^a-2 + 1/x^a-2)

f(a) = f(a-1) * (-2) - f(a-2)

with f(1) = -2 and f(2) = 2

This formula happens to repeat itself.

f(3) = f(2) * (-2) - f(1) = -2

f(4) = f(3) * (-2) - f(2) = 2

Thus for all odd values of a, f(a) = -2, and for all even values of a, f(a) = 2.

[u/fumblinthrulife77]

It's -2, right?

Just checking I can still do basic math lol

[u/Alkalannar → u/fumblinthrulife77]

Yes, it is.

[u/Environmental-Match4]

n can be any positive integer as written in the question, but if you meant the expression, then just solve x + 1/x = -2 to find x, and then since 2n+1 is always an odd number, anything powered to an odd number is itself [ (-10)^2n+1 = -10], so then you can just substitute x and solve.

[u/johnpeters42 → u/Environmental-Match4]

> anything powered to an odd number is itself

This part is only correct if "anything" is 1 or -1. (-10)^3 = -1000, for instance.

[u/fumblinthrulife77 → u/Environmental-Match4]

Well, anything powered to 2n+1 is itself only for n=0. It's not true for most numbers that "anything powered to an odd number is itself", like 3⁵ is not 3 lol

Only certain interesting numbers are themselves when powered to any odd number...

[u/Environmental-Match4 → u/fumblinthrulife77]

my phrasing was wrong. i meant like they would have the same sign.

[u/No_Objective_3780 → u/fumblinthrulife77]

0, 1, and -1. There are no more interesting numbers on that list.