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Question Number 118759 by ZiYangLee last updated on 19/Oct/20

The first three terms in the binomial expansion (p−q)^m , in ascending order of q, are denoted by a,b and c respectively. Show that (b^2 /(ac))=((2m)/(m−1))

$$\mathrm{The}\:\mathrm{first}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{expansion} \\ $$$$\left({p}−{q}\right)^{{m}} \:,\:\mathrm{in}\:\mathrm{ascending}\:\mathrm{order}\:\mathrm{of}\:{q},\:\mathrm{are}\:\mathrm{denoted} \\ $$$$\mathrm{by}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{respectively}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\frac{{b}^{\mathrm{2}} }{{ac}}=\frac{\mathrm{2}{m}}{{m}−\mathrm{1}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

a=p^m b=mp^(m−1) .(−q) c= ((m(m−1))/2)p^(m−2) .q^2 (b^2 /(ac))=((m^2 p^(2m−2) .q^2 )/(p^m .((m(m−1))/2)p^(m−2) .q^2 )) = ((2m)/(m−1)) proved

$$\mathrm{a}=\mathrm{p}^{\mathrm{m}} \\ $$$$\mathrm{b}=\mathrm{mp}^{\mathrm{m}−\mathrm{1}} .\left(−\mathrm{q}\right) \\ $$$$\mathrm{c}=\:\frac{\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}}\mathrm{p}^{\mathrm{m}−\mathrm{2}} .\mathrm{q}^{\mathrm{2}} \\ $$$$\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{ac}}=\frac{\mathrm{m}^{\mathrm{2}} \mathrm{p}^{\mathrm{2m}−\mathrm{2}} .\mathrm{q}^{\mathrm{2}} }{\mathrm{p}^{\mathrm{m}} .\frac{\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}}\mathrm{p}^{\mathrm{m}−\mathrm{2}} .\mathrm{q}^{\mathrm{2}} }\:=\:\frac{\mathrm{2m}}{\mathrm{m}−\mathrm{1}}\:\mathrm{proved} \\ $$

Commented by ZiYangLee last updated on 19/Oct/20