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Question Number 172688 by mr W last updated on 30/Jun/22

$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \:\:{right}? \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} \:\:{right}? \\ $$

Commented by Rasheed.Sindhi last updated on 30/Jun/22

$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \: \\ $$$$\:\:\:\:\:\:{False} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} \: \\ $$$$\:\:\:\:\:{True} \\ $$

Commented by mr W last updated on 30/Jun/22

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Commented by mahdipoor last updated on 30/Jun/22

$$\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +…+{a}^{\mathrm{3}} =\left(\frac{{a}\left({a}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{2}+…+{a}\right)^{\mathrm{2}} \\ $$

Commented by mr W last updated on 30/Jun/22

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