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Question Number 123175 by mathocean1 last updated on 23/Nov/20
Show by recurrence that  ∀ n ∈ N, Σ_(k=0 ) ^(n−1) q^k =((q^n −1)/(q−1))
$${Show}\:{by}\:{recurrence}\:{that} \\ $$$$\forall\:{n}\:\in\:\mathbb{N},\:\sum_{{k}=\mathrm{0}\:} ^{{n}−\mathrm{1}} {q}^{{k}} =\frac{{q}^{{n}} −\mathrm{1}}{{q}−\mathrm{1}}\: \\ $$
Answered by PNL last updated on 24/Nov/20
for n=1  Σ_(k=0) ^0 q^k =q^0 =((q^1 −1)/(q−1))=1 , so it′s true for n=1    for n>1, n∈N, let′s suppose this property true and  let′s prouve it for n+1    we have for n+1 :  Σ_(k=0) ^n q^k =Σ_(k=0) ^(n−1) q^k + q^n =((q^n −1)/(q−1))+ q^n =((q^n −1+q^(n+1) −q^n )/(q−1))=((q^(n+1) −1)/(q−1))   so it′s true for n+1    therefore Σ_(k=0) ^(n−1) q^k =((q^k −1)/(q−1))  for n∈N
$${for}\:{n}=\mathrm{1} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{0}} {\sum}}{q}^{{k}} ={q}^{\mathrm{0}} =\frac{{q}^{\mathrm{1}} −\mathrm{1}}{{q}−\mathrm{1}}=\mathrm{1}\:,\:{so}\:{it}'{s}\:{true}\:{for}\:{n}=\mathrm{1} \\ $$$$ \\ $$$${for}\:{n}>\mathrm{1},\:{n}\in\mathbb{N},\:{let}'{s}\:{suppose}\:{this}\:{property}\:{true}\:{and}\:\:{let}'{s}\:{prouve}\:{it}\:{for}\:{n}+\mathrm{1} \\ $$$$ \\ $$$${we}\:{have}\:{for}\:{n}+\mathrm{1}\:: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{q}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{q}^{{k}} +\:{q}^{{n}} =\frac{{q}^{{n}} −\mathrm{1}}{{q}−\mathrm{1}}+\:{q}^{{n}} =\frac{{q}^{{n}} −\mathrm{1}+{q}^{{n}+\mathrm{1}} −{q}^{{n}} }{{q}−\mathrm{1}}=\frac{{q}^{{n}+\mathrm{1}} −\mathrm{1}}{{q}−\mathrm{1}}\: \\ $$$${so}\:\boldsymbol{{it}}'\boldsymbol{{s}}\:\boldsymbol{{true}}\:\boldsymbol{{for}}\:\boldsymbol{{n}}+\mathrm{1} \\ $$$$ \\ $$$${therefore}\:\underset{{k}=\mathrm{0}} {\overset{\boldsymbol{{n}}−\mathrm{1}} {\sum}}\boldsymbol{{q}}^{\boldsymbol{{k}}} =\frac{\boldsymbol{{q}}^{\boldsymbol{{k}}} −\mathrm{1}}{\boldsymbol{{q}}−\mathrm{1}}\:\:\boldsymbol{{for}}\:\boldsymbol{{n}}\in\mathbb{N} \\ $$

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