Question

In the expansion of ( x^3 - 2/x^2 ) ^10 , find the coefficient of 1/x^5​

Answer 1

Answer:

240

Step-by-step explanation:

We need to find the coeffeicent of the binomial expansion of

$( {x}^{3} - 2 {x}^{ - 2} ) {}^{10}$

Note that

$- 2 {x}^{ - 2} = - \frac{2}{ {x}^{2} }$

The binomial theorem states that

$(x + y) {}^{n} = x {}^{n} y {}^{0} + \binom{n}{1} x {}^{n - 1} y + \binom{n}{2} x {}^{n - 2} y {}^{2} ....... + x {}^{0} y {}^{n} ( \binom{n}{n} )$

Using this, we let expand our series

$( {x}^{3} - 2 {x}^{ - 2} ) {}^{10} = x {}^{30} + \binom{10}{1} ( {x}^{27} 2 {x}^{ - 2} ) + \binom{10}{2} {x}^{24} 2x {}^{ - 4} + \binom{10}{3} {x}^{21} 2x {}^{ - 6} + \binom{10}{4} {x}^{18} 2x { }^{ - 8} + \binom{10}{5} x {}^{15} 2x {}^{ - 10} + \binom{10}{6} x {}^{12}2 x {}^{ - 12} + \binom{10}{7} x {}^{ 9} 2x {}^{ - 14} + \binom{10}{8} x {}^{ 6} 2x {}^{ - 16} + \binom{10}{9} ( {x}^{3} )2x {}^{ - 18} + 2x {}^{ - 20}$

$\frac{1}{ {x}^{5} } = x {}^{ - 5}$

So what term in the series eqaul x^-5.

That term is the 10 choose 7 term.

$\binom{10}{7} {x}^{9} 2x {}^{ - 14}$

Because

$= \binom{10}{7} 2x {}^{ - 14} {x}^{9} = \binom{10}{7} 2 {x}^{ - 5}$

So we need to compute 10 choose 7.

That equals

10!/3!(7!)= 10×9×8/6= 720/6=120.

So we get

$120(2) {x}^{ - 5}$

$240 {x}^{ - 5}$

So the coeffceint u

is 240