If the value of 1+23+63^2+103^3+ is , and _0.25(13+13^2+13^3+) is , and L=^, then L^2 is e…

Question

If the value of $1+\dfrac23+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\cdots\infty$ is $\alpha$, and $\log_{0.25}\left(\dfrac13+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\cdots\infty\right)$ is $\beta$, and $L=\alpha^{\beta}$, then $L^2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Let $\alpha=1+\dfrac23+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\cdots$. Using the subtraction technique ($\alpha-\tfrac{\alpha}{3}$), the numerators' second differences vanish, giving

$$\frac{2\alpha}{3} = 1+\frac13+\frac{4}{3^2}+\frac{4}{3^3}+\cdots = \frac43\cdot\frac{1}{1-1/3} = 2 \implies \alpha=3.$$

$$\frac13+\frac{1}{3^2}+\cdots = \frac{1/3}{1-1/3}=\frac12, \qquad \beta = \log_{0.25}\frac12 = \log_{1/4}\frac12 = \frac12.$$

$$L = \alpha^\beta = 3^{1/2} \implies L^2 = 3.$$

Solutions · 1