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Question Number 118937 by harckinwunmy last updated on 20/Oct/20

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{rule}$$$$\mathrm{for}\mathrm{factorising}{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^{\mathrm{n}}?$$

Answered by mathmax by abdo last updated on 20/Oct/20

$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{developping}$$$${(\mathrm{a}+\mathrm{b}+\mathrm{c})}^{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(\mathrm{a}+\mathrm{b})}^{\mathrm{k}}{\mathrm{c}}^{\mathrm{n}-\mathrm{k}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\left(\sum _{\mathrm{p}=0}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{k}}^{\mathrm{p}}{\mathrm{a}}^{\mathrm{p}}{\mathrm{b}}^{\mathrm{k}-\mathrm{p}}\right){\mathrm{c}}^{\mathrm{n}+\mathrm{k}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{n}}\sum _{\mathrm{p}=0}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{k}}^{\mathrm{p}}{\mathrm{a}}^{\mathrm{p}}{\mathrm{b}}^{\mathrm{k}-\mathrm{p}}{\mathrm{c}}^{\mathrm{n}+\mathrm{k}}$$

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