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

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

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

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

Commented by ZiYangLee last updated on 19/Oct/20

$$b={mp}^{m-1}\cdot (-q)$$

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

$$\mathrm{yes}$$

Commented by ZiYangLee last updated on 22/Oct/20

$$★$$

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