Sum of coefficients on polynomial

Hello.

There's this problem in my algebra book:

Problem 101. Imagine that the polynomial $\displaystyle (1+2x)^{200}$ is converted to the standard form (the sum of powers of x with numerical coefficients). What is the sum of all the coefficients?

My solution. The solution of this problem is $\displaystyle 2^{200}$ I think, am I right?

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**DIOGYK** My solution. The solution of this problem is $\displaystyle 2^{200}$ I think, am I right?

Hint: the sum of coefficients of a polynomial P is P(1).

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**emakarov** Hint: the sum of coefficients of a polynomial P is P(1).

So it's $\displaystyle 3^{200}$?

Re: Sum of coefficients on polynomial

Re: Sum of coefficients on polynomial

I don't know, because when I learned pascal's triangle, since it's about coefficients of polynomials, sum of the coefficients were always $\displaystyle 2^n$, isn't that right? or I think I'm confused maybe.

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**DIOGYK** I don't know, because when I learned pascal's triangle, since it's about coefficients of polynomials, sum of the coefficients were always $\displaystyle 2^n$, isn't that right?

Binomial coefficients can be used to find the sum of polynomial coefficients, but each polynomial requires its own approach. You don't claim that the sum of coefficients of *any* polynomial is $\displaystyle 2^n$, do you?

In this particular problem, it is not necessary to use binomial coefficients. Do you see that the sum of coefficients of a polynomial P is P(1)?

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**DIOGYK** I don't know, because when I learned pascal's triangle, since it's about coefficients of polynomials, sum of the coefficients were always $\displaystyle 2^n$, isn't that right? or I think I'm confused maybe.

Pascal's triangle works if you're expanding something like $\displaystyle (1+x)^n$. If you have something else, like $\displaystyle (1+2x)^n$, you need both binomial coefficients and the $\displaystyle 2^k$ factor. Basically, you can't say that the sum of coefficients is always $\displaystyle 2^n$.

On the other hand, P(1) gives you the sum of coefficients.

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**richard1234** Pascal's triangle works if you're expanding something like $\displaystyle (1+x)^n$. If you have something else, like $\displaystyle (1+2x)^n$, you need both binomial coefficients and the $\displaystyle 2^k$ factor. Basically, you can't say that the sum of coefficients is always $\displaystyle 2^n$.

On the other hand, P(1) gives you the sum of coefficients.

I understand, Thanks for help.

Quote:

Originally Posted by

**emakarov** Binomial coefficients can be used to find the sum of polynomial coefficients, but each polynomial requires its own approach. You don't claim that the sum of coefficients of *any* polynomial is $\displaystyle 2^n$, do you?

In this particular problem, it is not necessary to use binomial coefficients. Do you see that the sum of coefficients of a polynomial P is P(1)?

Thank you. I understand it now. And yes, I see that sum of coefficients of a polynomial is P(1). Thanks for help.

Also, is it okay if I create a new thread on every problem that I will be unable to solve?

Re: Sum of coefficients on polynomial

Quote:

Originally Posted by

**DIOGYK** Also, is it okay if I create a new thread on every problem that I will be unable to solve?

Yes; this is preferred way.