Question 1127761
ascending powers of {{{x}}} the expansion of {{{(2-x/2)^6 }}}

 (the sort starts from the smallest or lowest value)


{{{(2-x/2)^6 = 6C0* (2)^(6-0)(-x/2)^0 + 6C1* (2)^(6-1)*(-x/2)^1 + 6C2* (2)^(6-2)*(-x/2)^2

+ 6C3 *(2)^(6-3)*(-x/2)^3 + 6C4* (2)^(6-4)*(-x/2)^4 + 6C5* (2)^(6-5)*(-x/2)^5

+ 6C6* (2)^(6-6)*(-x/2)^6 }}}

since {{{6C0 =1}}},{{{6C1 =6}}},{{{6C2 =15}}},{{{6C3 =20}}},{{{6C4 =15}}},{{{6C5 =6}}}, and {{{6C6 =1}}}, we have


{{{(2-x/2)^6 =1*2^6 *1+6*2^5*(-x/2)+15*2^4(-x/2)^2+20*2^3(-x/2)^3+15*2^2(-x/2)^4+6*2(-x/2)^5+1*2^0*(-x/2)^6 }}}


{{{(2-x/2)^6 =64+192*(-x/2)+240(-x/2)^2+160(-x/2)^3+60(-x/2)^4+12(-x/2)^5+(-x/2)^6 }}}

{{{(2-x/2)^6 =64-192x/2+240x^2/4-160x^3/8+60x^4/16-12x^5/32+x^6 /64}}}


{{{(2-x/2)^6 =64-96x+60x^2-20x^3+15x^4/4-3x^5/8+x^6 /64}}}