How do you find the binomial expansion for ((x-(2/x^2))^9?

Dec 26, 2017

According to the binomial theorem, the expansion of an expression ( in general ) is:
${\left(x + a\right)}^{n} = {\text{^nC_0 x^n a^0 + ""^nC_1 x^(n-1) a^1 + ""^nC_2 x^(n-2) a^2+..........+ ""^nC_r x^(n-r) a^r + .......... + ""^nC_(n-1) x^1 a^(n-1) + }}^{n} {C}_{n} {x}^{0} {a}^{n}$

Hmmm.........

That does not look nice........

Well, let's see what those strange notations actually mean:
""^nC_r=(n!)/[{(n-r)!}r!]

r! = r* (r-1) * (r-2) * ........ * 2 * 1

So now, I hope all is cleared.
Thus, using the binomial theorem,
${\left(x + n\right)}^{9} = {\text{^9C_0 x^9 n^0 + ""^9C_1 x^8 n^1 + ""^9C_2 x^7 n^2 + ""^9C_0 x^6 n^3 + ""^9C_4 x^5 n^4 + ""^9C_5 x^4 n^5 + ""^9C_6 x^3 n^6 + ""^9C_7 x^2 n^7 + ""^9C_8 x^1 n^8 + }}^{9} {C}_{9} {x}^{0} {n}^{9} = {x}^{9} + 9 {x}^{8} {n}^{1} + 36 {x}^{7} {n}^{2} + 84 {x}^{6} {n}^{3} + 126 {x}^{5} {n}^{4} + 126 {x}^{4} {n}^{5} + 84 {x}^{3} {n}^{6} + 36 {x}^{2} {n}^{7} + 9 {x}^{1} {n}^{8} + {n}^{9}$

So, we can now solve your question:
${\left(x - \frac{2}{x} ^ 2\right)}^{9}$
 = x^9 - 9x^8*2/x^2 + 36 x^7 * 4/x^4 - 84 x^6*8/x^8 + 126 x^5*16/x^16 - 126 x^4*32/x^32 + 84 x^3*64/x^64 - 36 x^2*128/x^128 + 9x*256/x^256 - 512/x^512
$= {x}^{9} - 18 {x}^{6} + 144 {x}^{3} - \frac{672}{x} ^ 2 + \frac{2016}{x} ^ 11 - \frac{4032}{x} ^ 28 + \frac{5376}{x} ^ 61 - \frac{4608}{x} ^ 126 + \frac{2304}{x} ^ 247 - \frac{512}{x} ^ 512$

WHEW!!! that was something!!! Typing in the matter was so time - consuming; dude!