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Question Number 32707 by Rio Mike last updated on 31/Mar/18

$$1)\mathrm{find}\mathrm{u}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{u}\mathrm{if}$$$$\sum _{\mathrm{n}=1}^{4}2u.2^{n-1}=64$$$$2)\mathrm{find}\mathrm{k}\mathrm{if}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{k}.{\left(\frac{1}{3}\right)}^{\mathrm{n}-1}=\frac{2}{3}$$

Answered by MJS last updated on 31/Mar/18

$$\underset{n=1}{\sum ^4}2u\times 2^{n-1}=2u\underset{n=1}{\sum ^4}{2}^{n-1}=$$$$=2u(1+2+4+8)=30u$$$$u=\frac{64}{30}=\frac{32}{15}$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}k\times {\left(\frac{1}{3}\right)}^{n-1}=k\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{3^n}=k\times \frac{3}{2}$$$$\frac{3k}{2}=\frac{2}{3}$$$$k=\frac{4}{9}$$

Commented by NECx last updated on 02/Apr/18

$$howdid\underset{n=0}{\sum ^{\propto }}k\times {\left(1/3\right)}^{n-1}become3k/2$$$$???$$

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