Question

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

Answer 1

240

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