Question Number 163675 by alcohol last updated on 09/Jan/22
$${how}\:{do}\:{we}\:{find}\:{the}\:{sum}\:{of}\:{the}\:{terms}\:{after}\: \\ $$$${the}\:{n}^{{th}} \:{term}\:{of}\:{a}\:{GP} \\ $$
Commented by mr W last updated on 09/Jan/22
$${make}\:{the}\:\left({n}+\mathrm{1}\right)^{{th}} \:{term}\:{to}\:{the}\:{first} \\ $$$${term}\:{and}\:{apply}\:{the}\:{formula}\:{for}\:{sum} \\ $$$${from}\:{first}\:{term}\:{to}\:\infty. \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{1}} {q}+{a}_{\mathrm{1}} {q}^{\mathrm{2}} +…+{a}_{\mathrm{1}} {q}^{{n}−\mathrm{1}} +{a}_{\mathrm{1}} {q}^{{n}} +{a}_{\mathrm{1}} {q}^{{n}+\mathrm{1}} +…=\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}} \\ $$$${a}_{\mathrm{1}} {q}^{{n}} +{a}_{\mathrm{1}} {q}^{{n}+\mathrm{1}} +…=\frac{{a}_{\mathrm{1}} {q}^{{n}} }{\mathrm{1}−{q}} \\ $$$$ \\ $$$${or} \\ $$$${sum}\:{from}\:{first}\:{to}\:{n}^{{th}} \:{term}:\:\frac{{a}_{\mathrm{1}} \left(\mathrm{1}−{q}^{{n}} \right)}{\mathrm{1}−{q}} \\ $$$${sum}\:{from}\:{first}\:{term}\:{to}\:\infty\::\:\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}} \\ $$$${sum}\:{from}\:\left({n}+\mathrm{1}\right)^{{tb}} \:{term}\:{to}\:\infty\::\:\frac{{a}_{\mathrm{1}} }{\mathrm{1}−{q}}−\frac{{a}_{\mathrm{1}} \left(\mathrm{1}−{q}^{{n}} \right)}{\mathrm{1}−{q}}=\frac{{a}_{\mathrm{1}} {q}^{{n}} }{\mathrm{1}−{q}} \\ $$