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 10 - 2019 - Paper 2

To demonstrate this identity, we first rewrite the expression using logarithm properties:

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

This is a telescoping series. When expanded, many terms will cancel out:

• The first few and the last terms will remain:

$= \left( \log_{5}(3) - \log_{5}(2) \right) + \left( \log_{5}(4) - \log_{5}(3) \right) + \ldots + \left( \log_{5}(50) - \log_{5}(49) \right)$

Notice that most intermediate logs cancel:

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

Now evaluate the expression:

$= \log_{5}\left(\frac{50}{2}\right) = \log_{5}(25)$

Since $25$ can be expressed as $5^{2}$, we find:

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

Thus, we have shown that:

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