Question 1059424
Look at the general binomial expansion,
{{{(x+y)^15=sum(C(15,x)*x^(15-n)*y^n,n=1,15)}}}
1st term {{{n=0}}}
{{{C(15,0)*x^(15-0)*y^0=1*x^15*1=x^15}}}
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2nd term {{{n=2}}}
{{{C(15,1)*x^(15-1)*y^1=15*x^14*y=15x^14y}}}
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3rd term {{{n=3}}}
{{{C(15,2)*x^(15-2)*y^2=105*x^13*y^2}}}
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Remaining terms,
{{{455*x^12*y^3}}}
{{{1365*x^11*y^4}}}
{{{3003*x^10*y^5}}}
{{{5005*x^9*y^6}}}
{{{6435*x^8*y^7}}}
{{{6435*x^7*y^8}}}
{{{5005*x^6*y^9}}}
{{{3003*x^5*y^10}}}
{{{1365*x^4*y^11}}}
{{{455*x^3*y^12}}}
{{{105*x^2*y^13}}}
{{{15*x*y^14}}}
{{{y^15}}}
Now substitute,
{{{x=a}}} and {{{y=-2b}}}
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{{{x^15=a^15}}}
{{{15x^14y=15a^14(-2b)=-30a^14b}}}
{{{105*x^13*y^2=105*a^13* (-2b) ^2=420a^13b)=^2}}}
{{{455*x^12*y^3=455*a^12*(-2b)^3=    -3640a^12b^3}}}
{{{1365*x^11*y^4=1365*a^11*(-2b)^4=  21840a^11b^4}}}
{{{3003*x^10*y^5=3003*a^10*(-2b)^5=  -96096a^10b^5}}}
{{{5005*x^9*y^6=5005*a^9*(-2b)^6=    320320a^9b^6}}}
{{{6435*x^8*y^7=6435*a^8*(-2b)^7=    -823680a^8b^7}}}
{{{6435*x^7*y^8=6435*a^7*(-2b)^8=    1647360a^7b^8}}}
{{{5005*x^6*y^9=5005*a^6*(-2b)^9=    -2562560a^6b^9}}}
{{{3003*x^5*y^1=3003*a^5*(-2b)^1=    3075072a^5b^10}}}
{{{1365*x^4*y^11=1365*a^4*(-2b)^11=  -2795520a^4b^11}}}
{{{455*x^3*y^12=455*a^3*(-2b)^12=    1863680a^3b^12}}}
{{{105*x^2*y^13=105*a^2*(-2b)^13=    -860160a^2b^13}}}
{{{15*x*y^14=15*a*(-2b)^14=          245760a^1b^14}}}
{{{y^15=(-2b)^15=-32768b^15}}}
So then just sum all the terms to get the binomial expansion.