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Write-k-0-n-2-3-k-in-terms-of-n-Where-n-N-




Question Number 135642 by Raxreedoroid last updated on 14/Mar/21
Write Σ_(k=0) ^(n−2) 3^k  in terms of n. Where n∈N.
$$\mathrm{Write}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\mathrm{3}^{{k}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}.\:\mathrm{Where}\:{n}\in\mathbb{N}. \\ $$
Answered by Ñï= last updated on 14/Mar/21
Σ_(k=0) ^(n−2) 3^k =1+3+3^2 +...+3^(n−2) =((3^(n−1) −1)/(3−1))=(1/2)(3^(n−1) −1)
$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\mathrm{3}^{{k}} =\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +…+\mathrm{3}^{{n}−\mathrm{2}} =\frac{\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}\right) \\ $$
Answered by Olaf last updated on 14/Mar/21
Σ_(k=0) ^(n−2) 3^k  = 3^0 ×((1−3^(n−1) )/(1−3)) = (1/2)(3^(n−1) −1)
$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\mathrm{3}^{{k}} \:=\:\mathrm{3}^{\mathrm{0}} ×\frac{\mathrm{1}−\mathrm{3}^{{n}−\mathrm{1}} }{\mathrm{1}−\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}\right) \\ $$

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