math question

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[graceful summit]

Since each term is a power of x, the sum of all coefficients is just this function evaluated at x = 1.

And to find the arithmetic mean, recall how many terms (a + b)^n has.

[subtle wind]

it has n + 1 terms?

am i right

or wrong

[graceful summit]

Right, yeah.

[subtle wind]

soo ((-1)^n)/n is 0.2

(-1)^n = 0.2n but n cant be nagative so i cant say that it is 5

[graceful summit]

Why divided by n?

There are (n + 1) terms, not n.

[subtle wind]

ohh right

yes

(-1)^n = 0.2n + 0.2 so n is 4
x^(3*2) / x^(-4) = x^2
C(5, 2) = 10
10x4 = 40

is my solution right?

and i have 1 more question

[graceful summit]

When typing a *, put a \ before it, otherwise it's used for formatting.

[subtle wind]

ohh okk

[graceful summit]

Also, how did you find the term with x^2?

[rose elk]

Why is it $\frac{(-1)^n}{n+1}=0.2$?

[thorn cairn]

Satyam

[graceful summit]

Sum of terms is (-1)^n, and there are (n + 1) of them.

[subtle wind]

c(n, r).(x)^(3.(n-r)).x^(-2r).(-2)^(r)

[rose elk]

Did you mean $\binom{n}{0}-2\binom{n}{1}+2^2\binom{n}{2}-2^3\binom{n}{3}+\dots=(-1)^n$?

[thorn cairn]

Satyam

[graceful summit]

Ah, then yes.
The term is C(4, r) x^(3(4 - r)) (-2/x^2)^(r) = C(4, r) (-2)^r x^(12 - 5r). We want x^2, so 12 - 5r = 2, so r = 2.

[subtle wind]

ohh i typed 5 instead of 4 yes

[graceful summit]

Sum of coefficients, yeah, sorry.

[rose elk]

Since the arithmetic mean is given and we're dealing with coefficients?

[subtle wind]

so 6.4 = 24

C

[rose elk]

?

[graceful summit]

What?
We are given that the average of coefficients is 1/5. Their sum is (1 - 2)^n = (-1)^n, and there are (n + 1) of them, so the average is (-1)^n/(n + 1).

[graceful summit]

Yeah.

[subtle wind]

i have 1 more question

[rose elk]

Maybe I'm confused but how sum of the coefficients is (-1)^n?

[subtle wind]

An is an arithmetic sequence
A2 = 2A1 + 1
A6 + A22 = 34
find A7

[rose elk]

You can proceed.

[graceful summit]

Each term is a power function, so determining the sum of coefficients is as easy as substituting x = 1.

[graceful summit]

Well, you have a(n) = a(1) + (n - 1)d, so you can find a(1) and d from the system.

[subtle wind]

i found that An = (2^n).A1 + (2^n -1)

but it comes out too big

[rose elk]

Will it work for any binomial? like substituting x=1?

[graceful summit]

That's not an arithmetic progression.

[graceful summit]

Not quite: both terms need to be power functions, and you have to check that you don't get the same powers in different terms.

Here one of the terms has a positive exponent and another has a negative one, so that's sufficient.

[rose elk]

Can you give a example of such binomial where we can't just substitute x=1 to get sum of coefficients?

[subtle wind]

oh i think i solved it is it C ?

[graceful summit]

Hm... On second though, I think it'll be possible for any expression of the form (ax^m + bx^n)^k, as long as m and n aren't equal. After all, the term in this case is C(k, r) a^(k - r) b^r x^(mk - mr) x^(nr) = C(k, r) a^(k - r) b^r x^(mk - (m - n)r), and mk - (m - n)r is a monotone function when r changes from 0 to k.

[graceful summit]

Yes.

[rose elk]

Okay thanks, I didn't know that.

[subtle wind]

thanks for helping

[fluid linden]

@rose elk @graceful summit next time please discuss this in #math-discussion or #helper-staff

[graceful summit]

You're welcome!

Ah, yeah, sorry.

[rose elk]

Sorry.

[rose elk]

DId you get the answer?

[subtle wind]

yes

[rose elk]

close the post using +close.

[subtle wind]

okk