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Hint: We have been given${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$, solve it for$ r=1,2,3$. Then use$\cos \theta +i\sin \theta ={{e}^{i\theta }}$to convert and then use the formula for the sum of infinite G.P, which is${{s}_{\infty }}=\dfrac{a}{1-r}$. Solve it, you will get the answer.

Complete step-by-step answer:

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

For example, the sequence $2,6,18,....$ is a geometric progression with a common ratio $3$.

Similarly, $10,5,2.5,......$ is a geometric sequence with common ratio $\dfrac{1}{2}$.

Examples of a geometric sequence are powers of a fixed number ${{r}^{k}}$, such as ${{2}^{k}}$ and ${{3}^{k}}$.

The general form of a geometric sequence is,

$a,ar,a{{r}^{2}},....$

The ${{n}^{th}}$ term of a geometric sequence with initial value$a={{a}_{0}}$ and the common ratio $r$ is given by,

${{a}_{n}}=a{{r}^{n-1}}$

Such a geometric sequence also follows the recursive relation.

${{a}_{n}}={{a}_{^{n-1}}}r$ for every integer $n\ge 1$.

Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio.

The common ratio of a geometric sequence may be negative, resulting in an alternating sequence, with numbers alternating between positive and negative.

The behavior of a geometric sequence depends on the value of the common ratio.

If the common ratio is:

$*$ Positive, the terms will all be the same sign as the initial term.

$*$ Negative, the terms will alternate between positive and negative.

$*$ Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).

$*$ 1, the progression is a constant sequence.

$*$ Between âˆ’1 and 1 but not zero, there will be exponential decay towards zero.

$*$ âˆ’1, the progression is an alternating sequence.

$*$ Less than âˆ’1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign.

Now we have been given that,

${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$

We know $\cos \theta +i\sin \theta ={{e}^{i\theta }}$.

Now put $r=1$.

${{X}_{1}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$

Now put $r=2$.

${{X}_{2}}=\cos \dfrac{\pi }{{{2}^{2}}}+i\sin \dfrac{\pi }{{{2}^{2}}}=\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}={{e}^{i\dfrac{\pi }{4}}}$

Also, now put $r=3$.

${{X}_{3}}=\cos \dfrac{\pi }{{{2}^{3}}}+i\sin \dfrac{\pi }{{{2}^{3}}}=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}={{e}^{i\dfrac{\pi }{8}}}$

So for\[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}\]

Here we can see that\[\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.....\]is a series of G.P.

Because we can see the common ratio $r=\dfrac{1}{2}$.

Finding the sum of terms in a geometric progression is easily obtained by applying the formula :

${{n}^{th}}$partial sum of a geometric sequence,

${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$

So for$\infty $,

${{S}_{\infty }}=\dfrac{a}{1-r}$

Here the first term$=a=\dfrac{1}{2}$ and $r=\dfrac{1}{2}$.

So using the formula we get,

${{S}_{\infty }}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}=1$

So \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}={{e}^{\pi i(1)}}={{e}^{\pi i}}\]

So we get,

\[{{e}^{\pi i}}=\cos \pi +i\sin \pi \]

We know that $\cos \pi =-1$ and $\sin \pi =0$.

Substituting these values in the above equation, we get,

\[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{\pi i}}=\cos \pi +i\sin \pi =-1+0=-1\]

So we get the final answer \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty =-1\].

Hence proved.

Note: Read the question in a careful manner. You should be familiar with the concept of G.P. You must know the sum of $n$ terms which are ${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$ and also, for $\infty $ which is ${{S}_{\infty }}=\dfrac{a}{1-r}$. Donâ€™t make a mistake while simplifying. Most students make silly mistakes.

Complete step-by-step answer:

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

For example, the sequence $2,6,18,....$ is a geometric progression with a common ratio $3$.

Similarly, $10,5,2.5,......$ is a geometric sequence with common ratio $\dfrac{1}{2}$.

Examples of a geometric sequence are powers of a fixed number ${{r}^{k}}$, such as ${{2}^{k}}$ and ${{3}^{k}}$.

The general form of a geometric sequence is,

$a,ar,a{{r}^{2}},....$

The ${{n}^{th}}$ term of a geometric sequence with initial value$a={{a}_{0}}$ and the common ratio $r$ is given by,

${{a}_{n}}=a{{r}^{n-1}}$

Such a geometric sequence also follows the recursive relation.

${{a}_{n}}={{a}_{^{n-1}}}r$ for every integer $n\ge 1$.

Generally, to check whether a given sequence is geometric, one simply checks whether successive entries in the sequence all have the same ratio.

The common ratio of a geometric sequence may be negative, resulting in an alternating sequence, with numbers alternating between positive and negative.

The behavior of a geometric sequence depends on the value of the common ratio.

If the common ratio is:

$*$ Positive, the terms will all be the same sign as the initial term.

$*$ Negative, the terms will alternate between positive and negative.

$*$ Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).

$*$ 1, the progression is a constant sequence.

$*$ Between âˆ’1 and 1 but not zero, there will be exponential decay towards zero.

$*$ âˆ’1, the progression is an alternating sequence.

$*$ Less than âˆ’1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign.

Now we have been given that,

${{X}_{r}}=\cos \dfrac{\pi }{{{2}^{r}}}+i\sin \dfrac{\pi }{{{2}^{r}}}$

We know $\cos \theta +i\sin \theta ={{e}^{i\theta }}$.

Now put $r=1$.

${{X}_{1}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}={{e}^{i\dfrac{\pi }{2}}}$

Now put $r=2$.

${{X}_{2}}=\cos \dfrac{\pi }{{{2}^{2}}}+i\sin \dfrac{\pi }{{{2}^{2}}}=\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}={{e}^{i\dfrac{\pi }{4}}}$

Also, now put $r=3$.

${{X}_{3}}=\cos \dfrac{\pi }{{{2}^{3}}}+i\sin \dfrac{\pi }{{{2}^{3}}}=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}={{e}^{i\dfrac{\pi }{8}}}$

So for\[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}\]

Here we can see that\[\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.....\]is a series of G.P.

Because we can see the common ratio $r=\dfrac{1}{2}$.

Finding the sum of terms in a geometric progression is easily obtained by applying the formula :

${{n}^{th}}$partial sum of a geometric sequence,

${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$

So for$\infty $,

${{S}_{\infty }}=\dfrac{a}{1-r}$

Here the first term$=a=\dfrac{1}{2}$ and $r=\dfrac{1}{2}$.

So using the formula we get,

${{S}_{\infty }}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}=1$

So \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{i\dfrac{\pi }{2}}}.{{e}^{i\dfrac{\pi }{4}}}.{{e}^{i\dfrac{\pi }{8}}}......={{e}^{\pi i\left( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+... \right)}}={{e}^{\pi i(1)}}={{e}^{\pi i}}\]

So we get,

\[{{e}^{\pi i}}=\cos \pi +i\sin \pi \]

We know that $\cos \pi =-1$ and $\sin \pi =0$.

Substituting these values in the above equation, we get,

\[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty ={{e}^{\pi i}}=\cos \pi +i\sin \pi =-1+0=-1\]

So we get the final answer \[{{X}_{1}}{{X}_{2}}{{X}_{3}}.......\infty =-1\].

Hence proved.

Note: Read the question in a careful manner. You should be familiar with the concept of G.P. You must know the sum of $n$ terms which are ${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$ and also, for $\infty $ which is ${{S}_{\infty }}=\dfrac{a}{1-r}$. Donâ€™t make a mistake while simplifying. Most students make silly mistakes.

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