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Question Number 34707 by mondodotto@gmail.com last updated on 10/May/18

$$\mathbf{given}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{first}\mathbf{n}\mathbf{terms}$$$$\mathbf{of}\mathbf{an}\mathbf{A}.\mathbf{P}\mathbf{is}{\mathbf{n}}^2\mathbf{the}\mathbf{sum}\mathbf{of}$$$$\mathbf{of}\mathbf{the}\mathbf{first}2\mathbf{n}\mathbf{terms}\mathbf{of}\mathbf{the}$$$$\mathbf{same}\mathbf{A}.\mathbf{P}\mathbf{is}{\mathbf{n}}^2+\mathbf{n}.$$$$\mathbf{show}\mathbf{that}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{first}$$$$4\mathbf{n}\mathbf{terms}\mathbf{is}4{\mathbf{n}}^2-8\mathbf{n}+4.$$

Commented by Rasheed.Sindhi last updated on 10/May/18

$$\mathrm{If}\mathrm{sum}\mathrm{of}\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}={\mathrm{n}}^2$$$$\mathrm{then}$$$$\mathrm{the}\mathrm{sum}\mathrm{of}2\mathrm{n}\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{AP}={\left(2\mathrm{n}\right)}^2={4\mathrm{n}}^2$$$$\ne {\mathrm{n}}^2+\mathrm{n}$$$$\mathrm{The}\mathrm{sum}\mathrm{of}4\mathrm{n}\mathrm{terms}={\left(4\mathrm{n}\right)}^2={16\mathrm{n}}^2$$$$\mathrm{The}\mathrm{sum}\mathrm{of}4\mathrm{n}\mathrm{terms}\ne 4{\mathbf{n}}^2-8\mathbf{n}+4$$

Commented by mondodotto@gmail.com last updated on 10/May/18

$$\mathrm{please}\mathrm{recheck}$$

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