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Question Number 174844 by daus last updated on 12/Aug/22

Commented by daus last updated on 12/Aug/22

pls help for D   a^k  + a^(k+2)  +..... n^(th)  term

$${pls}\:{help}\:{for}\:{D}\: \\ $$$${a}^{{k}} \:+\:{a}^{{k}+\mathrm{2}} \:+.....\:{n}^{{th}} \:{term} \\ $$

Answered by Rasheed.Sindhi last updated on 12/Aug/22

D.  a^k  + a^(k+2)  +..... n^(th)  term  a=a^k , r=a^(k+2) /a^k =a^(k+2−k) =a^2   S_n =((a(r^n −1))/(r−1))=((a^k ( (a^2 )^n −1 ))/(a^2 −1))  =a^k ( (a^2 )^(n−1) +(a^2 )^(n−2) +...+a^2 +1)  =a^k ( a^(2n−2) +a^(2n−4) +...+a^2 +1)

$${D}. \\ $$$${a}^{{k}} \:+\:{a}^{{k}+\mathrm{2}} \:+.....\:{n}^{{th}} \:{term} \\ $$$$\boldsymbol{{a}}={a}^{{k}} ,\:\boldsymbol{{r}}={a}^{{k}+\mathrm{2}} /{a}^{{k}} ={a}^{{k}+\mathrm{2}−{k}} ={a}^{\mathrm{2}} \\ $$$$\boldsymbol{{S}}_{\boldsymbol{{n}}} =\frac{\boldsymbol{{a}}\left(\boldsymbol{{r}}^{\boldsymbol{{n}}} −\mathrm{1}\right)}{\boldsymbol{{r}}−\mathrm{1}}=\frac{{a}^{{k}} \left(\:\left({a}^{\mathrm{2}} \right)^{{n}} −\mathrm{1}\:\right)}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$={a}^{{k}} \left(\:\left({a}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} +\left({a}^{\mathrm{2}} \right)^{{n}−\mathrm{2}} +...+{a}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$={a}^{{k}} \left(\:{a}^{\mathrm{2}{n}−\mathrm{2}} +{a}^{\mathrm{2}{n}−\mathrm{4}} +...+{a}^{\mathrm{2}} +\mathrm{1}\right) \\ $$

Commented by daus last updated on 13/Aug/22

how to decide which formula to use because   i don know r>1 or r<1 ?

$${how}\:{to}\:{decide}\:{which}\:{formula}\:{to}\:{use}\:{because}\: \\ $$$${i}\:{don}\:{know}\:{r}>\mathrm{1}\:{or}\:{r}<\mathrm{1}\:? \\ $$

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