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Question Number 118790 by mathocean1 last updated on 19/Oct/20

$$\mathrm{Show}\mathrm{by}\mathrm{recurence}\mathrm{that}$$$${(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\times {\mathrm{a}}^{\mathrm{k}}\times {\mathrm{b}}^{\mathrm{n}-\mathrm{k}}$$

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