Find the coefficient of the term $x^2$ in $$\left(x+\frac 2x\right)^{4}$$
This is what I have done so far.
We know $a$ = $x$ and $b$ = $\frac2x$ and $n$ = 4
So what I have done is $4c2$ x $(x)^2$ x ($\frac2x$)$^2$
As we know that $^2$ + $^2$ = 4, and 4 is the $n$ value.
The answer is one though? What have I done wrong?
P.S. this was the tutorial that helped me.
On
The general formula you need to use is:
$$kx^2 = \binom{n}{r} \cdot a^{n-r} \cdot b^r$$
where $n = 4$, $a = x$, $b = \frac{2}{x}$, and $k$ is the co-efficient we're trying to find.
After substituting you get:
$$kx^2 = \binom{4}{r} \cdot x^{4-r} \cdot \left(\frac{2}{x}\right)^r$$
To find the value of $r$ we only need powers of $x$, because we know they must be equal:
$$x^2 = x^{4-r} \cdot x^{-r}$$ Therefore, $$2 = (4 - r) - r$$ $$r = 1$$
Finally, substitute $r$ with $1$ in the original equation:
$$kx^2 = \binom{4}{1} \cdot x^{4-1} \cdot \left(\frac{2}{x}\right)^1$$ $$kx^2 = (4) \cdot (x^3) \cdot \left(\frac{2}{x}\right)$$ $$kx^2 = 8x^2$$ $$k = 8$$
And that's your answer, the co-efficient of the term $x^2$ is $8$.
On
$$\left(x+\frac 2x\right)^{4} = \sum_{k=0}^4 \binom{4}{k} x^k \left(\frac{2}{x}\right)^{4-k} = \sum_{k=0}^4 \binom{4}{k} x^{2k-4} 2^{4-k}$$
If we want the coefficient of $x^2$, then $k=3$. Can you take it from here?
On
Your task is to find all ways of multiplying $x$ and $1/x$ four times in total so that the degree is $2$.
So this is the same as finding which combinations of $\pm 1\pm 1 \pm 1 \pm 1$ equals $2$.
But after a moment's thought, you see that there is only one solution up to permutations: $1+1+1-1$. This yield four possible ways.
Hence the coefficient is $4 \cdot 2=8$ (since each $1/x$ has a two in front).
In the original version, the exponent was $64$
The $r+1$th term will be $$\binom4rx^{4-r}\left(\dfrac2x\right)^r=\binom4rx^{4-2r}2^r$$
We need $4-2r=2$