Question

If I = √-1 a n € N , then in + in + 1 + in + 2 + in + 3 is equal to ​

Answer 1

$\rm\huge\red{\underline{\text{Proper question:-}}}$

If $\rm i^2=-1$ and $\rm n\in\mathbb{N}$, find the value of $\rm i^n+i^{n+1}+i^{n+2}+i^{n+3}$.

$\rm\Huge\red{\underline{\text{Idea:-}}}$

$\bold{Imaginary\ unit\ i}$

We know that $i^2=1$ which is a definition of $i$.

$\bold{Factorization}$

The HCF of each term is $i^n$.

We can group the common factors by polynomial identity.

Which is,

$\large\rightarrow\red{\boxed{\red{\bold{ak+bk=k(a+b)}}}}$

$\huge\red{\underline{\red{\text{Solution:-}}}}$

Given number is,

$\rm\rightarrow i^n+i^{n+1}+i^{n+2}+i^{n+3}$

The HCF of each term is,

$\rm\rightarrow i^n$

So by grouping the common factor,

$\rm\rightarrow i^n(i^0+i^1+i^2+i^3)$

If we simplify,

$\rm\rightarrow i^{n}(1+i-1-i)$

$\rm=i^{n}\cdot0$

$\rm=\red{\underline{\bold{0}}\tiny\text{//}}$

So, the answer is 0.

$\Huge\red{\underline{\red{\text{Learn more:-}}}}$

$\bold{Cycling\ powers\ of\ i}$

$\large\rightarrow\red{\boxed{\red{\bold{i^{4k}=1}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{i^{4k+1}=i}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{i^{4k+2}=-1}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{i^{4k+3}=-i}}}}$

$k$ belongs to integers.

$\bold{The\ cube\ roots\ of\ unity\ \omega}$

The roots of the equation $\rm x^3=1$ contain exactly one real and two imaginary solutions.

$\large\hookrightarrow\red{\boxed{\red{\bold{\omega=\dfrac{-1\pm\sqrt{3}i}{2}}}}}$

$\bold{Properties\ of\ \omega}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega^{3k}=1}}}}$

$\rm k$ belongs to integers.

$\large\rightarrow\red{\boxed{\red{\bold{\omega^2+\omega+1=0}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega+\bar{\omega}=-1}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega\bar{\omega}=1}}}}$

$\bold{Variations}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega^{3k+1}=\omega}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega^{3k+2}=\omega^2}}}}$

$\large\rightarrow\red{\boxed{\red{\bold{\omega=\dfrac{1}{\bar{\omega}}}}}}$

Answer 2

Answer:

.

• $\red{ \boxed{ {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = 0} }$

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Step-by-step explanation:

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$\hookrightarrow \text{ i }= \sqrt{ - 1} \: \: \: \text{ ...(given)}$

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We can say that,

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$\hookrightarrow { \text{ i }}^{2} = { - 1} \: \: \: ...(1)$

$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3}$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} + {\text{i}}^{\text{n}} \times {\text{i} }^{1} + {\text{i}}^{\text{n}} \times {\text{i} }^{2} + {\text{i}}^{\text{n} } \times {\text{i} }^{3}$

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${Taking} \: \: { \text{i}}^{ \text{n} } \: {common,}$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} (1 + {\text{i} } + {\text{i} }^{2} + {\text{i} }^{3})$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} (1 + {\text{i} } - 1 + {\text{i} }^{2 + 1})$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} (1 + {\text{i} } - 1 + {\text{i} }^{2 } \times\text{i} )$

.

$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} ( \cancel{1} + \sqrt{ - 1} - \cancel{1 } + {\text{i} }^{2 } \times\text{i} )$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} ( \sqrt{ - 1} + - 1 \times( \sqrt{ - 1} ) )$

.

$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} ( \sqrt{ - 1} - 1( \sqrt{ - 1} ) )$

.

$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} ( \cancel { \sqrt{ - 1} } - \cancel{1( \sqrt{ - 1} )} )$

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$\hookrightarrow {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = {\text{i}}^{\text{n}} (0)$

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$\hookrightarrow \red{ \boxed{ {\text{i}}^{\text{n}} + {\text{i}}^{\text{n} + 1} + {\text{i}}^{\text{n} + 2} + {\text{i}}^{\text{n} + 3} = 0} }$

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