8. (i) Find the value of $$\sum_{r=4}^{\infty} 20 \times \left(\frac{1}{2}\right)^{r}$$ (ii) Show that $$\sum_{n=1}^{48} \log_{5}\left(\frac{n+2}{n+1}\right) = 2$$ - Edexcel - A-Level Maths Pure - Question 11 - 2019 - Paper 2

Using the property of logarithms, we can rewrite the sum as:

$\sum_{n=1}^{48} \log_{5}\left(\frac{n+2}{n+1}\right) = \sum_{n=1}^{48} \left( \log_{5}(n+2) - \log_{5}(n+1) \right)$

This forms a telescoping series. Writing out the first few terms, we have:

• For $n=1$: $\log_{5}(3) - \log_{5}(2)$
• For $n=2$: $\log_{5}(4) - \log_{5}(3)$
• For $n=3$: $\log_{5}(5) - \log_{5}(4)$
• Continue this pattern until $n=48$: $\log_{5}(50) - \log_{5}(49)$

Notice how most of the terms cancel:

$= \left(\log_{5}(50) - \log_{5}(2)\right)$

This means:

$\sum_{n=1}^{48} \log_{5}\left(\frac{n+2}{n+1}\right) = \log_{5}(50) - \log_{5}(2) = \log_{5}\left(\frac{50}{2}\right) = \log_{5}(25)$

Since $25 = 5^{2}$, we have:

$\log_{5}(25) = 2$