Let (x+10)^50 + (x-10)^50 = a0 + a1x + a2x^2 + .... + a50x^50, for all x∈R then a2/a0 is equal to:

Given:-(x+10)50+(x𕒵0)50=a0+a1x+a2x2+...+a50x50To find:-a2a0=coefficient of x2Coefficient of x0As we know that, the general term in an expansion(a+b)nis given as,Tr+1=nCr(a)n−r(b)rNow,General term of(x+10)50-Here,a=x,b=10Tr+1=50Cr(x)50−r(10)rFor coefficient ofx2-50−r=2⇒r=48T48+1=50C48(x)50󔼸(10)48T49=50C48(10)48x2For coeficient ofx0-50−r=0⇒r=50T50+1=50C50(x)50󔼺(10)50T51=50C50(10)50x0Now,General term of(x+(𕒵0))50-Here,a=x,b=𕒵0Tr+1=50Cr(x)50−r(𕒵0)rFor coeficient ofx2-50−r=2⇒r=48T48+1=50C48(x)50󔼸(𕒵0)48T49=50C48(10)48x2For coeficient ofx0-50−r=0⇒r=50T50+1=50C50(x)50󔼺(𕒵0)50T51=50C50(10)50x0Now from the given expansion,a2=50C48(10)48+50C48(10)48=50C48((10)48+(10)48)a0=50C50(10)50+50C50(10)50=50C50((10)50+(10)50)Now,a2a0=50C48((10)48+(10)48)50C50((10)50+(10)50)As we know that,nCr=n!r!(n−r)!Therefore,a2a0=50×492×(10481050)⇒a2a0=494=12.25