Remember that #((n),(k)) = _nC_k = (n!)/((n-k)!k!)#
#(2x-3)^10 = _10C_0(2x)^10(-3)^0 + _10C_1(2x)^9(-3)^0 + _10C_2(2x)^8(-3)^2 + _10C_3(2x)^7(-3)^3 + _10C_4(2x)^6(-3)^4 + _10C_5(2x)^5(-3)^5 + _10C_6(2x)^4(-3)^6 + _10C_7(2x)^3(-3)^7 + _10C_8(2x)^2(-3)^8 + _10C_9(2x)^1(-3)^9 + _10C_10(2x)^0(-3)^10#
You can use Pascal's Triangle to find your combinations or the formula above. From Pascal's triangle #: _10C_0 = _10C_10 = 1;
_10C_1 = _10C_9 = 10;
_10C_2 = _10C_8 = 45;
_10C_3 = _10C_7 = 120;
_10C_4 = _10C_6 = 210;
_10C_5 = 252#
Substitute into the equation and simplify:
#(2x-3)^10 = 1024x^10 + 10(512)x^9(-3) + 45(256)x^8(9) + 120(128)x^7(-27) + 210(64)x^6(81) + 252(32)x^5(-243) + 210(16)x^4(729) + 120(8)x^3(-2187) + 45(4)x^2(6561) + 10(2)x(-19683) + 59049#
Simplify by multiplying the constants:
#(2x-3)^10 = 1024x^10 -15360x^9 + 103680x^8 - 414720x^7 + 1088640x^6 - 1959552x^5 + 2449440x^4 - 2099520x^3 + 1180980x^2 - 393660x + 59049#
