Photo AI

A geometric series is $a + ar + ar^2 + ...$ (a) Prove that the sum of the first $n$ terms of this series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Question icon

Question 2

A-geometric-series-is-$a-+-ar-+-ar^2-+-...$----(a)-Prove-that-the-sum-of-the-first-$n$-terms-of-this-series-is-given-by---$S_n-=-\frac{a(1---r^n)}{1---r}$-Edexcel-A-Level Maths Pure-Question 2-2006-Paper 2.png

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) Fi... 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 $S_n = \frac{a(1 - r^n)}{1 - r}$ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 2

Step 1

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

96%

114 rated

Answer

To prove this, we start with the sum of the first nn terms of a geometric series:

Sn=a+ar+ar2++arn1.S_n = a + ar + ar^2 + \ldots + ar^{n-1}.

If we multiply both sides by (1r)(1 - r):

(1r)Sn=a(1r)+ar(1r)+ar2(1r)++arn1(1r).(1 - r)S_n = a(1 - r) + ar(1 - r) + ar^2(1 - r) + \ldots + ar^{n-1}(1 - r).

This simplifies to:

(1r)Sn=aarn,(1 - r)S_n = a - ar^n,

Rearranging gives:

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

Step 2

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

99%

104 rated

Answer

Using the formula for the sum of a geometric series:

Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}

Here, a=100(21)=200a = 100(2^1) = 200, r=2r = 2, and n=10n = 10.

We can substitute these values into the formula:

S10=200(1210)12=200(2101)=200(10241)=200×1023=204600.S_{10} = \frac{200(1 - 2^{10})}{1 - 2} = 200(2^{10} - 1) = 200(1024 - 1) = 200 \times 1023 = 204600.

Step 3

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

96%

101 rated

Answer

For the series to converge, the common ratio must satisfy r<1|r| < 1.

In this case, the first term a=56a = \frac{5}{6} and the common ratio r=13r = \frac{1}{3}.

The sum to infinity is given by:

S=a1r=56113=5623=56×32=1512=54.S = \frac{a}{1 - r} = \frac{\frac{5}{6}}{1 - \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{2}{3}} = \frac{5}{6} \times \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.

98%

120 rated

Answer

The condition for convergence of an infinite geometric series is:

r<1.|r| < 1.

This means that the absolute value of the common ratio must be less than 1.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;