2.1 Consider the geometric series: 4 + 2 + 1 + \frac{1}{2} + .. - NSC Mathematics - Question 2 - 2024 - Paper 1

To calculate the sum of the infinite series, we use the formula for the sum of a converging geometric series:

$S_\infty = \frac{a}{1 - r}$

Substituting the values of $a$ and $r$:

$S_\infty = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 4 \times 2 = 8$

Thus, the sum of the series is $S_\infty = 8$.

First, we write the sum of a geometric series:

$\sum_{p=k}^{n} ar^{p} = a \frac{r^{n+1} - r^{k}}{r - 1}$

In our case, $a = 3^k$, $r = 3$, and $n = 10$. Thus:

$29 520 = 3^k \frac{3^{11} - 1}{3 - 1}$

Calculate $3^{11}$:

$3^{11} = 177147$

Now, substituting back:

$29 520 = 3^k \frac{177147 - 1}{2}$

This simplifies to:

$29 520 = 3^k \times 88573$

Next, we isolate $3^k$:

$3^k = \frac{29 520}{88573}$

Calculating this gives:

$3^k = 0.3333$

To find $k$, we recognize that:

$3^k=3^1\implies k = 1$.