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\displaystyle S = \sum_{r=2}^\infty \frac{r2^{r-2}}{3^{r-1}} = \frac{1}{3}\sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2}.\\[br]\text{Let } u_n = \sum_{r=n}^\infty \bigg( \frac{2}{3} \bigg)^{r-2} = 3\bigg( \frac{2}{3} \bigg)^{n-2}.\\[br]\text{Then } S = \frac{1}{3}(2u_2+u_3+u_4+u_5+\dots). \\[br]u_2 = 3\\[br]u_2+u_3+u_4+\dots = \frac{3}{1 - \frac{2}{3}} = 9.\\[br]\therefore S = \frac{1}{3}(3+9) = 4.
III/14
\text{Solving this equation we have }\int\dfrac{dV}{aK-bV}=\int\dfrac{dt}{K}\Rightarrow-\dfrac{1}{b}\ln(aK-bV)=\dfrac{1}{K}-\dfrac{1}{b}\lnaK \text{ since }V=0 \text{ at }t=0
\text{i.e. }\ln\left( \dfrac{aK-bV}{aK}\right)=-\dfrac{bt}{K}\RightarrowaK-bV=aK\text{exp}\left(-\dfrac{bt}{K}\right) \text{ or }V=\dfrac{aK}{b}\left(1-\text{e}^{-\frac{bt}{K}}\right)\text{ as required}
\text{so }\int\left(\dfrac{dV}{(bL-cK)V}}\right)=\int\dfrac{dt}{KL}\Rightarrow -\dfrac{1}{c}\ln{(bL-cK)V}=\dfrac{t}{KL}-\dfrac{1}{c}\ln{bL} \text{ since }V=0\text{ at }t=0
\text{By Newton's second law }Mg-T=\dfrac{d}{dt}(Mv)= \dotM v+M\dotv \text{ where }T \text{ is the tension at Q}
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