Binomial theorem - not an easy question

[renvox]

Hi guys, I'm Filip and as a 11th grade student I have a question about one mathematical problem. It says:
If the coefficient of x^k in the expansion of
(3+2x-x² )*(1+x)³⁴ is zero. Find the value of k.
I know it's something related with binomial theorem, but I don't really know how to start.
Thank you guys for your help!

 

[fzero]

First of all, you should attempt to factor (3+2x-x²). Then you should expand (1+x)ⁿ into series form. Multiply by the remaining factor and obtain an equation for the coefficient of the x^k term, a_k that will involve the binomial coefficients. Then try to solve a_k = 0.

 

[renvox]

Ok, so I factored (3+2x-x2) and it gave me -(x-3)(x+1).
But then i don't undersand what you meant by expanding (1+x)ⁿ into series form and the rest of what you have written. Could you help me please?


Filip

 

[fzero]

Well the binomial formula tells you that

(1+x)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k.

Are you familiar with that formula? Can you apply it here to find a series expression for

(3+2x-x^2)(1+x)^{34}?

 

[renvox]

Well the binomial formula tells you that

(1+x)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k.

Are you familiar with that formula? Can you apply it here to find a series expression for

(3+2x-x^2)(1+x)^{34}?

ok, it is:
<br /> (1+x)^34 = \sum_{k=1}^34 \begin{pmatrix} 34 \\ k \end{pmatrix} x^k.<br />
To be honest with you, i don't know if what I've done now is correct.
Besides, i don't know why i can't put blablabla to the power of 34 because it shows me to the power of 3 and 4 is somewhere don't (i mean this language latex)

 

[fzero]

ok, it is:
<br /> (1+x)^34 = \sum_{k=1}^34 \begin{pmatrix} 34 \\ k \end{pmatrix} x^k.<br />
To be honest with you, i don't know if what I've done now is correct.
Besides, i don't know why i can't put blablabla to the power of 34 because it shows me to the power of 3 and 4 is somewhere don't (i mean this language latex)

You can put the 34 in brackets, {34} to get the exponent to work out properly in LaTeX.

It will save you sometime to note that you were able to factor the quadratic term, so

<br /> (3+2x-x^2)(1+x)^{34} = - (x-3)(x+1)(1+x)^{34} = - (x-3)(1+x)^{35} = - (x-3) \sum_{k=1}^{35} \begin{pmatrix} 35 \\ k \end{pmatrix} x^k.<br />

The next step is to take the factor of (x-3) into the sum. You will need to play around with the index that the sum runs over to put it into the form

\sum_{k} a_k x^k + \cdots

in order to try to solve a_k=0.

 

[renvox]

What do you mean by taking the factor of (x-3) into the sum?
May you give me an example in order to understand what you mean?

 

[fzero]

What do you mean by taking the factor of (x-3) into the sum?
May you give me an example in order to understand what you mean?

Sure, I mean distribute the product over the terms of the sum. So, if we had

(x+2) \sum_{k=0}^n b_k x^k = \sum_{k=0}^n b_k ( x^{k+1} + 2 x^k) .

Note that we can rewrite the first term by setting s= k+1 as

\sum_{k=0}^n b_k x^{k+1} = \sum_{s=1}^{n+1} b_s x^s .

You'll need to make a similar transformation so that you can compare the coefficients of terms of the same power of x.