6 Calculate the value of the $$ n $$th term. a $$ \quad x(1+i) ; x(1+i)^{2} ; x(1+i)^{3} ; \ldots $$ b $$ x(1+i)^{-1} ; x(1+i)^{-2} ; x(1+i)^{-3} ; \ldots $$ c $$ an x ; \sin x ; \sin x \cos x ; \ldots $$ d $$ 1 ; \sin x ; 1-\cos ^{2} x ; \ldots $$

Question

6 Calculate the value of the $$ n $$th term. a $$ \quad x(1+i) ; x(1+i)^{2} ; x(1+i)^{3} ; \ldots $$ b $$ x(1+i)^{-1} ; x(1+i)^{-2} ; x(1+i)^{-3} ; \ldots $$ c $$ 	an x ; \sin x ; \sin x \cos x ; \ldots $$ d $$ 1 ; \sin x ; 1-\cos ^{2} x ; \ldots $$

UpStudy AI · Accepted Answer

Sure, let's determine the $$ n $$th term for each of the given sequences.

### a) Sequence: $$ x(1+i) ;\ x(1+i)^{2} ;\ x(1+i)^{3} ;\ \ldots $$

This is a **geometric sequence** where each term is obtained by multiplying the previous term by the common ratio $$ (1+i) $$.

- **First term ($$ n = 1 $$)**: $$ x(1+i) $$
- **Common ratio**: $$ r = (1+i) $$

**$$ n $$th term ($$ a_n $$)**:
$$
a_n = x(1+i)^n
$$

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### b) Sequence: $$ x(1+i)^{-1} ;\ x(1+i)^{-2} ;\ x(1+i)^{-3} ;\ \ldots $$

This is also a **geometric sequence**, but with a common ratio of $$ (1+i)^{-1} $$.

- **First term ($$ n = 1 $$)**: $$ x(1+i)^{-1} $$
- **Common ratio**: $$ r = (1+i)^{-1} $$

**$$ n $$th term ($$ a_n $$)**:
$$
a_n = x(1+i)^{-n}
$$

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### c) Sequence: $$ \tan x ;\ \sin x ;\ \sin x \cos x ;\ \ldots $$

Let's analyze the pattern:

1. **First term ($$ n = 1 $$)**: $$ \tan x = \frac{\sin x}{\cos x} $$
2. **Second term ($$ n = 2 $$)**: $$ \sin x $$
3. **Third term ($$ n = 3 $$)**: $$ \sin x \cos x $$

Observing the pattern, each term from the second onward multiplies by $$ \cos x $$:

- To transition from $$ \tan x $$ to $$ \sin x $$: Multiply by $$ \cos x $$.
- To transition from $$ \sin x $$ to $$ \sin x \cos x $$: Multiply by $$ \cos x $$.

Thus, the general form involves multiplying $$ \sin x $$ by $$ \cos x $$ raised to $$ (n-2) $$.

**$$ n $$th term ($$ a_n $$)**:
$$
a_n = \sin x \cdot (\cos x)^{n-2}
$$

*Alternatively*, recognizing that $$ \tan x = \sin x \cdot \cos^{-1} x $$:
$$
a_n = \sin x \cdot \cos^{n-2} x
$$

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### d) Sequence: $$ 1 ;\ \sin x ;\ 1 - \cos^{2}x ;\ \ldots $$

Let's identify the pattern:

1. **First term ($$ n = 1 $$)**: $$ 1 $$
2. **Second term ($$ n = 2 $$)**: $$ \sin x $$
3. **Third term ($$ n = 3 $$)**: $$ 1 - \cos^{2}x = \sin^{2}x $$

Observing the progression:

- Each term from the second onward is $$ \sin x $$ raised to the power $$ (n-1) $$.

**$$ n $$th term ($$ a_n $$)**:
$$
a_n = (\sin x)^{n-1}
$$

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**Summary of $$ n $$th Terms:**

a) $$ a_n = x(1+i)^n $$

b) $$ a_n = x(1+i)^{-n} $$

c) $$ a_n = \sin x \cdot (\cos x)^{n-2} $$

d) $$ a_n = (\sin x)^{n-1} $$
The $$ n $$th terms are: a) $$ a_n = x(1+i)^n $$ b) $$ a_n = x(1+i)^{-n} $$ c) $$ a_n = \sin x \cdot (\cos x)^{n-2} $$ d) $$ a_n = (\sin x)^{n-1} $$

6 Calculate the value of the \( n \)th term. a \( \quad x(1+i) ; x(1+i)^{2} ; x(1+i)^{3} ; \ldots \) b \( x(1+i)^{-1} ; x(1+i)^{-2} ; x(1+i)^{-3} ; \ldots \) c \( \tan x ; \sin x ; \sin x \cos x ; \ldots \) d \( 1 ; \sin x ; 1-\cos ^{2} x ; \ldots \)

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