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

1 Answer
Dec 26, 2017

According to the binomial theorem, the expansion of an expression ( in general ) is:
#(x+a)^n=""^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) + ""^nC_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,
#(x+n)^9=""^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 + ""^9C_9 x^0 n^9 = x^9 + 9x^8n^1 + 36 x^7n^2 + 84x^6n^3 + 126 x^5n^4 + 126 x^4n^5 + 84 x^3n^6 + 36 x^2n^7 + 9 x^1n^8 + n^9#

So, we can now solve your question:
#(x-2/x^2)^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 - 18x^6 + 144 x^3 - 672/x^2 + 2016/x^11 - 4032/x^28 + 5376/x^61 - 4608/x^126 + 2304/x^247 - 512/x^512#

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