A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find $\sum_{k=1}^{10} 100(2^k)$. (c) Find the sum to infinity of the geometric series $\frac{5}{6} + \frac{5}{18} + \frac{5}{54} + ...$. (d) State the condition for an infinite geometric series with common ratio $r$ to be convergent.

A-Level Edexcel Maths Pure Integration: A geometric series is $a + ar +

A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find $\sum_{k=1}^{10} 100(2^k)$ - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Question 1

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A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find ... show full transcript

Worked Solution & Example Answer:A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find $\sum_{k=1}^{10} 100(2^k)$ - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Step 1

Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$

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Answer

To derive the formula for the sum of the first $n$ terms of a geometric series, we start by expressing the series:

S_n = a + ar + ar^2 + ... + ar^{n-1}.

Multiplying both sides by $r$ , we have:

rS_n = ar + ar^2 + ar^3 + ... + ar^n.

Now, subtracting the second equation from the first gives:

S_n - rS_n = a - ar^n.

Factoring out $S_n$ results in:

S_n(1 - r) = a(1 - r^n).

Dividing both sides by $(1 - r)$ (assuming $r \neq 1$ ) gives:

S_n = \frac{a(1 - r^n)}{1 - r}.

Step 2

Find $\sum_{k=1}^{10} 100(2^k)$

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Answer

To find the sum:

\sum_{k=1}^{10} 100(2^k) = 100 \sum_{k=1}^{10} (2^k)

Using the sum formula for a geometric series:

\sum_{k=0}^{n} ar^k = a \frac{(1 - r^{n+1})}{1 - r}

where $a = 2$ , $r = 2$ , and $n = 10 - 1 = 9$ (starting from $2^1$ ), we find:

\sum_{k=1}^{10} 2^k = 2(1 - 2^{10}) / (1 - 2) = 2(1 - 1024) / -1 = 2(1023) = 2046.

Thus,

100 \times 2046 = 204600.

Step 3

Find the sum to infinity of the geometric series $\frac{5}{6} + \frac{5}{18} + \frac{5}{54} + ...$

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Answer

The sum to infinity for a geometric series is given by:

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

where $a$ is the first term and $r$ is the common ratio.

Here, the first term $a = \frac{5}{6}$ and the common ratio can be found by:

r = \frac{\frac{5}{18}}{\frac{5}{6}} = \frac{5}{18} \cdot \frac{6}{5} = \frac{6}{18} = \frac{1}{3}.

Now, applying the formula:

S_\infty = \frac{\frac{5}{6}}{1 - \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{2}{3}} = \frac{5}{6} \cdot \frac{3}{2} = \frac{15}{12} = \frac{5}{4}.

Step 4

State the condition for an infinite geometric series with common ratio $r$ to be convergent.

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Answer

An infinite geometric series converges if the absolute value of the common ratio $r$ is less than 1, or mathematically:

|r| < 1.

A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find $\sum_{k=1}^{10} 100(2^k)$ - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Worked Solution & Example Answer:A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ (b) Find $\sum_{k=1}^{10} 100(2^k)$ - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$

Find $\sum_{k=1}^{10} 100(2^k)$

Find the sum to infinity of the geometric series $\frac{5}{6} + \frac{5}{18} + \frac{5}{54} + ...$

State the condition for an infinite geometric series with common ratio $r$ to be convergent.

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