(a) Calculate, to four decimal places, the first ten terms of the sequence: $a_n = 1 + \left(-\frac{2}{5}\right)^n.$

$n$	$a_n$
1	$\boxed{?}$
2	$\boxed{?}$
3	$\boxed{?}$
4	$\boxed{?}$
5	$\boxed{?}$
6	$\boxed{?}$
7	$\boxed{?}$
8	$\boxed{?}$
9	$\boxed{?}$
10	$\boxed{?}$

(b) Does the sequence appear to have a limit?

• Yes
• No

(c) If so, calculate it. If not, enter DNE.

$\boxed{?}$

Neetesh Kumar | December 29, 2024

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This is the solution to Math 1c
Assignment: 11.1 Question Number 6
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Step-by-step solution:

The given sequence is:

$a_n = 1 + \left(-\frac{2}{5}\right)^n.$

Step 1: Calculate the first ten terms of the sequence:

For each $n$, substitute the value of $n$ into the formula $a_n$ and evaluate to four decimal places:

1. For $n = 1$:

   $a_1 = 1 + \left(-\frac{2}{5}\right)^1 = 1 - \frac{2}{5} = 1 - 0.4 = 0.6$

2. For $n = 2$:

   $a_2 = 1 + \left(-\frac{2}{5}\right)^2 = 1 + \left(\frac{4}{25}\right) = 1 + 0.16 = 1.16$

3. For $n = 3$:

   $a_3 = 1 + \left(-\frac{2}{5}\right)^3 = 1 - \frac{8}{125} = 1 - 0.064 = 0.936$

4. For $n = 4$:

   $a_4 = 1 + \left(-\frac{2}{5}\right)^4 = 1 + \frac{16}{625} = 1 + 0.0256 = 1.0256.$

5. For $n = 5$:

   $a_5 = 1 + \left(-\frac{2}{5}\right)^5 = 1 - \frac{32}{3125} = 1 - 0.01024 = 0.98976$

6. For $n = 6$:

   $a_6 = 1 + \left(-\frac{2}{5}\right)^6 = 1 + \frac{64}{15625} = 1 + 0.004096 = 1.004096$

7. For $n = 7$:

   $a_7 = 1 + \left(-\frac{2}{5}\right)^7 = 1 - \frac{128}{78125} = 1 - 0.0016384 = 0.9983616$

8. For $n = 8$:

   $a_8 = 1 + \left(-\frac{2}{5}\right)^8 = 1 + \frac{256}{390625} = 1 + 0.00065536 = 1.0007$

9. For $n = 9$:

   $a_9 = 1 + \left(-\frac{2}{5}\right)^9 = 1 - \frac{512}{1953125} = 1 - 0.000262144 = 0.9997$

10. For $n = 10$:

    $a_{10} = 1 + \left(-\frac{2}{5}\right)^{10} = 1 + \frac{1024}{9765625} = 1 + 0.0001048576 = 1.0001048$

Thus, the first ten terms are:

$n$	$a_n$
1	0.6
2	1.16
3	0.936
4	1.0256
5	0.98976
6	1.004096
7	0.9983616
8	1.0007
9	0.9997
10	1.0001048

Step 2: Determine if the sequence has a limit:

As $n$ increases, the term $\left(-\frac{2}{5}\right)^n$ approaches $0$ because $-\frac{2}{5}$ has an absolute value less than $1$:

$\displaystyle\lim_{n \to \infty} \left(-\frac{2}{5}\right)^n = 0.$

Thus, the sequence $a_n$ approaches:

$\displaystyle\lim_{n \to \infty} a_n = 1 + 0 = 1.$

Final Answer:

(a)

$n$	$a_n$
1	$\boxed{0.6}$
2	$\boxed{1.16}$
3	$\boxed{0.936}$
4	$\boxed{1.0256}$
5	$\boxed{0.98976}$
6	$\boxed{1.004096}$
7	$\boxed{0.9983616}$
8	$\boxed{1.0007}$
9	$\boxed{0.9997}$
10	$\boxed{1.0001048}$

(b) $\boxed{\text{Yes}}$

(c) $\boxed{1}$

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