# Question 8:

## Part (i) – Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=4}^{\infty} 20\times\left(\frac{1}{2}\right)^r = 20\left(\frac{1}{2}\right)^4 + 20\left(\frac{1}{2}\right)^5 + 20\left(\frac{1}{2}\right)^6 + ...$ | | |
| $= \frac{20(\frac{1}{2})^4}{1-\frac{1}{2}}$ | M1 | Applies $\frac{a}{1-r}$ for their $r$ where $-1 <$ their $r < 1$ and their value for $a$ |
| Complete strategy applying $\frac{20(\frac{1}{2})^4}{1-\frac{1}{2}}$ | M1 | Finds infinite sum using complete strategy |
| $= 2.5$ | A1 | 2.5 o.e. |

## Part (i) – Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=4}^{\infty} 20\times\left(\frac{1}{2}\right)^r = \sum_{r=1}^{\infty} 20\times\left(\frac{1}{2}\right)^r - \sum_{r=1}^{3} 20\times\left(\frac{1}{2}\right)^r$ | | |
| $= \frac{10}{1-\frac{1}{2}} - (10+5+2.5)$ or $= \frac{10}{1-\frac{1}{2}} - \frac{10(1-(\frac{1}{2})^3)}{1-\frac{1}{2}}$ | M1 | Applies $\frac{a}{1-r}$ for their $r$ and value for $a$ |
| Fully correct strategy | M1 | Finds infinite sum by completely correct strategy |
| $\{= 20 - 17.5\} = 2.5$ | A1 | 2.5 o.e. |

## Part (i) – Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=4}^{\infty} 20\times\left(\frac{1}{2}\right)^r = \sum_{r=0}^{\infty} 20\times\left(\frac{1}{2}\right)^r - \sum_{r=0}^{3} 20\times\left(\frac{1}{2}\right)^r$ | | |
| $= \frac{20}{1-\frac{1}{2}} - (20+10+5+2.5)$ or $= \frac{20}{1-\frac{1}{2}} - \frac{20(1-(\frac{1}{2})^4)}{1-\frac{1}{2}}$ | M1 | Applies $\frac{a}{1-r}$ for their $r$ and value for $a$ |
| Fully correct strategy | M1 | Finds infinite sum by completely correct strategy |
| $\{= 40 - 37.5\} = 2.5$ | A1 | 2.5 o.e. |

**Note:** Give M1 M1 A1 for correct answer of 2.5 from no working in (i)

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## Part (ii) – Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_5\left(\frac{3}{2}\right) + \log_5\left(\frac{4}{3}\right) + ...... + \log_5\left(\frac{50}{49}\right) = \log_5\left(\frac{3}{2}\times\frac{4}{3}\times...\times\frac{50}{49}\right)$ | M1 | Some evidence of applying the addition law of logarithms as part of a valid proof |
| Writing at least three terms including either first two and last, or first and last two | M1 | Begins to solve the problem by writing/combining at least three terms |
| $= \log_5\left(\frac{50}{2}\right)$ or $\log_5(25) = 2$ * | A1* | Correct proof leading to answer of 2 |

**Note:** The 2nd M1 can be gained by writing any of:
- listing $\log_5\left(\frac{3}{2}\right), \log_5\left(\frac{4}{3}\right), \log_5\left(\frac{50}{49}\right)$ or $\log_5\left(\frac{3}{2}\right), \log_5\left(\frac{49}{48}\right), \log_5\left(\frac{50}{49}\right)$
- $\log_5\left(\frac{3}{2}\right) + \log_5\left(\frac{4}{3}\right) + ...... + \log_5\left(\frac{50}{49}\right)$
- $\log_5\left(\frac{3}{2}\times\frac{4}{3}\times...\times\frac{50}{49}\right)$ **{this will also gain the 1st M1 mark}**

**Note:** Do not allow 2nd M1 if $\log_5\left(\frac{3}{2}\right), \log_5\left(\frac{4}{3}\right)$ are listed and $\log_5\left(\frac{50}{49}\right)$ is used for first time in applying formula $S_{48} = \frac{48}{2}\left(\log_5\left(\frac{3}{2}\right) + \log_5\left(\frac{50}{49}\right)\right)$

**Note – Listing all 48 terms:** Give M0 M1 A0 for $\log_5\left(\frac{3}{2}\right) + \log_5\left(\frac{4}{3}\right) + ... + \log_5\left(\frac{50}{49}\right) = 2$ {lists all terms}; Give M0 M0 A0 for $0.2519...+ 0.1787...+ 0.1386...+......+0.0125... = 2$ {all terms in decimals}

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## Part (ii) – Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{n=1}^{48}\log_5\left(\frac{n+2}{n+1}\right) = \sum_{n=1}^{48}(\log_5(n+2) - \log_5(n+1))$ | M1 | Uses subtraction law of logarithms |
| $= (\log_5 3 + \log_5 4 + ...... + \log_5 50) - (\log_5 2 + \log_5 3 + ...... + \log_5 49)$ | M1 | Begins to solve by writing at least three terms for each of $\log_5(n+2)$ and $\log_5(n+1)$ including either first two and last term, or first and last two terms |
| $= \log_5 50 - \log_5 2$ or $\log_5\left(\frac{50}{2}\right)$ or $\log_5(25) = 2$* | A1* | Correct proof leading to answer of 2 |

**Note:** The base of 5 can be omitted for M marks in part (ii), but must be included in final line.

**Note:** Give M1 M0 A0 (1st M) for implied use of subtraction law for $\sum_{n=1}^{48}\log_5\left(\frac{n+2}{n+1}\right) = 91.8237... - 89.8237... = 2$

**Note:** Give M1 M1 A1 for:
$$\sum_{n=1}^{48}\log_5\left(\frac{n+2}{n+1}\right) = \sum_{n=1}^{48}(\log_5(n+2)-\log_5(n+1))$$
$$= \log_5(3\times4\times......\times50) - \log_5(2\times3\times......\times49)$$
$$= \log_5\left(\frac{50!}{2}\right) - \log_5(49!) \text{ or } = \log_5(25\times49!) - \log_5(49!)$$
$$= \log_5 25 = 2$$

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