Question

Find number of terms

Answer 1

\[\huge (x^2 + 1 + \frac{ 1 }{ x^2 })^n\]

Answer 2

answer is 2n +1

Answer 3

so is it a square of something to make it like a binomial so it has 2n+1 terms

Answer 4

@kainui @Directrix

Answer 5

@geerky42

Answer 6

try to expand and see what happens

Answer 7

expand till nth exponent lol

Answer 8

3^n

Answer 9

i don't want to die expanading a binomial

Answer 10

tri*nomial

Answer 11

you could treat it as a binomial though

Answer 12

answer is 2n+1

Answer 13

So what? You can treat anything as anynomial.

(3x+2-1+1-1+1)^n is not considered a "hexanomial" lol

Answer 14

Are you sure it's 2n+1 because I think you're wrong.

Answer 15

the book think's the answer is 2n+1 not me

Answer 16

well 3n is one of the options

Answer 17

thinks*

Answer 18

Well, you're right. But I'm just saying that to force you to defend yourself against an _almost correct_ answer.

Answer 19

but i am not understanding why 2n+1 is correct or 3n is correct

Answer 20

You have x^2 times 1/x^2 as a product. That cancels to 1 and becomes part of the constant term.

Answer 21

@Kainui is wrong, 2n+1 is correct. Try to expand to n=2, n=3, n=4 and you'll see why

Answer 22

Just because it is true for first few n, it doesn't mean it is true for all n...

Answer 23

For n = 1 to 5, 2n + 1 works.

Answer 24

@geerky42 You can prove that easily, I say that because it helps you see the algebraic pattern

Answer 25

\[\huge \left(\begin{matrix}n-2 \\ 2\end{matrix}\right)\]
how much does this evaluatet to

Answer 26

i got this from a formula

Answer 27

Well, you prove it and help OP, the hell are you waiting for? @AeroSmith

Answer 28

@Kainui should be demoted to level 1 for his obvious lack of intellect of solving elementary basic algebra. @No.name 's answer 2n+1 is correct for any n

Answer 29

same goes for @geerky42, I hope not all level 99s are as useless

Answer 30

>mocking 99 SS users
>not doing anything

Guess you belong to our boat.

Answer 31

why 2n+1 is correct

Answer 32

i mean how do we get it , please explain to me

Answer 33

@No.name because if you look at the product, you have x^2 and 1/x^2. This gives 1. That get brought into the same constant term as 1*1 in the product.

Answer 34

@No.name Hint, the x^2 and 1/x^2 cancels, the 1 simply leave the original term that multiplies it as it is.

Answer 35

from this point expanding it a few more it'll become very obvious

Answer 36

A binomial expansion raised to the nth power has 2^n terms. This formula counts like terms of the expansion separately.

It seems to me that a trinomial expansion to the nth power would have 3^n terms although that is just a conjecture at this stage.

Answer 37

(2n+1 is true because first few n works.)

patterns always hold, my retrice

Answer 38

well it has n+1 terms @Directrix , if i am not wrong

Answer 39

like (x+y)^2 has 3 terms , 2+1

Answer 40

The sum of binomial coefficients raised to the nth power has the sum 2^n
like in (x+y)^2 has the sum 2^n , 2^2 = 4

Answer 41

>>like (x+y)^2 has 3 terms

The expansion of (x + y)^2 has 4 terms. The expansion term count is done BEFORE combination of like terms.

Answer 42

@No.name, you can also easily prove it using mathematical induction

Answer 43

okay , thanks evr1

Answer 44

\[\left(x^2 + 1 + \frac{ 1 }{ x^2 }\right)^n = -\left(1 - (x+\frac{1}{x})^2\right)^n\]

say \(x+\frac{1}{x} =A \)

\[-\left(1-A\right)^n\]

would you agree there will be \(n+1\) terms in this expansion ?

Answer 45

yes definately

Answer 46

except for constant term, every term is a power of A yes ?

Answer 47

yes

Answer 48

\[-(1-A)^n = -(1-A+A^2 - A^3 + \cdots + (-A)^n)\]

Answer 49

nvm it only complicates :/

Answer 50

no problem

Answer 51

@AeroSmith good idea http://meta.wikimedia.org/wiki/Cunningham%27s_Law

Answer 52

@kainui two things are wrong in this convo: your statement "Are you sure it's 2n+1 because I think you're wrong.", and your level.