Find the value of the coefficient of
1/x
in the expansion of (2x −
1/x )^5
.

Question

asadkarim7 · Answer

i need explanation

amistre64 · Answer

it looks like a binomial expansion, but the question itself seems a bit skewed

amistre64 · Answer

$${n\choose r} a^r\ b^{1-r}$$
i think is the structure of it

amistre64 · Answer

1/x is only alone in the last term of the expansion

and I think I got the exponents backwards above

$${n\choose r} a^{n-r}\ b^{r}$$
that looks better

$${5\choose 5} 2x^{5-5}\ (-1/x)^{5}$$

amistre64 · Answer

(5 5) = 1
2x^0 = 1

(-1/x)^5 = -1 * 1/x

mr.math · Answer

The formula is
$$(z+y)^n=\sum_{k=0}^n\left(\begin{matrix}n \ k\end{matrix}\right) x^{n-k}y^k$$
Now just substitute directly in the formula with $n=5$ and $k=1$ to find the coefficient of $\frac{1}{x}$. Obviously $y=-\frac{1}{x}$ and $z=2x$.

mr.math · Answer

You need only to evaluate that term, you don't have to find all terms (i.e the term at k=1).

amistre64 · Answer

{n \choose k} is simpler to type up

amistre64 · Answer

$${n \choose k}$$

mr.math · Answer

$${n\choose k} \text{   Nice! Thanks :) }$$

amistre64 · Answer

;)