The coefficient of x square for the expansion is x 2 2

The coefficient of the #x^2#-term is #80#.

First, we need to find the sixth row of Pascal's Triangle to determine the coefficient of the non-simplified terms of the expansion of the binomial expression. Remember that Pascal's Triangle begins with #1# as the first row and as the first and last entry of every other row. The middle terms of each row are obtained by adding the two terms from the row above.

                     1
                    / \
                  1    1
                 / \  / \
               1    2   1
              / \  / \  / \
             1   3   3   1
           /  \  / \  / \ / \
          1    4   6   4   1
        /  \  / \  / \  / \  / \
       1    5   10  10  5  1

Remember that in this case, #(x + 2)^5#, #a = x# and #b = 2#. The binomial theorem tells us that in the expansion, the terms will follow this pattern:

#a^5b^0 + a^4b^1 + a^3b^2 + a^2b^3 + a^1b^4 + a^0b^5#

Combining that with the coefficients from Pascal's Truangle gives us this expansion of #(x + 2)^5#:

#1(x^5)(2^0) + 5(x^4)(2^1) + 10(x^3)(2^2) + 10(x^2)(2^3) + 5(x^1)(2^4) + 1(x^0)(2^5)

The fourth term is the #x^2#-term, so we need to simplify it:
#10(x^2)(2^3) = 10(x^2)(8) = 80x^2

So, the coefficient of the #x^2#-term is #80#.

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I was given this question as a practise assignment and I am unsure of my answer.

The coefficient of $x^2$ in the expansion of $(x+\frac{1}{ax})^8$ is 7. Find the possible value of $a$.

I did $(x+\frac{1}{ax})^8$ = $(x (1+\frac{1}{ax^2}))^8$

Given my answer, does that mean that $a=7$?

Edit: Thank you for everyone's help! I think I understand it a little bit now. I will have to review this question again and practise more.

asked Mar 16, 2021 at 22:25

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hint

$$(x+\frac{1}{ax})^2=x^2+\frac{1}{a^2x^2}+\frac{2}{a}$$

So, you just need to find the coefficient of $ X $ in the expansion

$$(X+\frac{1}{a^2X}+\frac{2}{a})^4$$

answered Mar 16, 2021 at 22:30

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Two options:

1. With the binomial theorem: the expansion of $(a+b)^n$ is $$(a+b)^n = \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2+\cdots + \binom{n}{n-1}a^1b^{n-1} + \binom{n}{n}a^0b^n.$$ Set $a=x$, $b=\frac{1}{ax}$, $n=8$, and figure out which of the terms is the $x^2$ term. This will give you an expression involving $a$, which you can then solve for $a$.

2. With derivatives. Rewrite the binomial as $$\left(x + \frac{1}{ax}\right) = \frac{1}{x}\left(x^2 + \frac{1}{a}\right).$$ Therefore, $$\left(x + \frac{1}{ax}\right)^8 = \frac{1}{x^8}\left(x^2+\frac{1}{a}\right)^8.$$ The coefficient of $x^2$ will be the coefficient of $x^{10}$ in $(x^2+\frac{1}{a})^{8}$.

This can be found using derivatives: if $p(x)$ is a polynomial, then the constant term of $p^{(k)}(x)$ is $k!$ times the coefficient of $x^k$ in $p(x)$. So $p^{(k)}(0)$ will give you $k!$ times the coefficient of $x^k$ in $p(x)$. Replace $x^2$ with $y$, and look for the coefficient of $y^5$ in the expansion of $(y+\frac{1}{a})^8$ to find the coefficient of $x^{10}$ in the original, and from there you can get the one for $x^2$ in the original expression. This will give you an expression involving $a$, which you can then solve for $a$.

answered Mar 16, 2021 at 22:40

Arturo MagidinArturo Magidin

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Write the expression as ${1 \over a^8 x^8} (1+a x^2)^8$ and find the coefficient of $x^{10}$ in $(1+a x^2)^8$.

Use the binomial theorem to compute the coefficient of $x^{10}$ in $(1+a x^2)^8$. Hint: It is the 5th coefficient.

Call this expression $E$ (it will be a formula involving $a$ and some numbers). Then the coefficient of $x^{2}$ in the original expression is ${1 \over a^8}E$.

Then solve the expression ${1 \over a^8}E = 7$ for $a$. (You will need to replace $E$ by the expression you computed first before solving.)

answered Mar 16, 2021 at 22:34

copper.hatcopper.hat

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You want to solve

$${8 \choose k}x^{8-k}\left(\frac{1}{ax^2}\right)^k=7x^2$$

This can be expressed in the form

$${8 \choose k}\cdot\frac{1}{a^k}x^{8-3k}=7x^2$$

Solving $8-3k=2$ gives $k=2$. Solving $\displaystyle{8\choose2}\cdot\dfrac{1}{a^2}=7x^2$ gives $a^2=4$.

However, $-2$ turns out to be extraneous, so $2$ is the only solution.

answered Mar 16, 2021 at 22:57

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