a) If z = + i , find the value of z^n + 1z^n by using De moivre's theorem. [2] b) In a group of 12 students, 8 are boys and remaining girls. In how many ways can 5 students be selected for quiz…

a) If $z=\cos \theta +i\sin \theta$, find the value of $z^n+\frac{1}{z^n}$ by using De Moivre's theorem.

Solution:

Given: $$z=\cos \theta +i\sin \theta$$

By De Moivre's theorem, we know that: $$z^n=(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$$

Now, we need to find $\frac{1}{z^n}$. Since $z=\cos \theta +i\sin \theta$, its complex conjugate is $\overline{z}=\cos \theta -i\sin \theta$. We also know that: $$z\cdot \overline{z}=(\cos \theta +i\sin \theta )(\cos \theta -i\sin \theta )={\cos }^2\theta +{\sin }^2\theta =1$$

Thus, $\frac{1}{z}=\overline{z}=\cos \theta -i\sin \theta$. Similarly, $$\frac{1}{z^n}=(\cos \theta -i\sin \theta {)}^n=\cos (n\theta )-i\sin (n\theta )$$

Now, we compute $z^n+\frac{1}{z^n}$: $$z^n+\frac{1}{z^n}=[\cos (n\theta )+i\sin (n\theta )]+[\cos (n\theta )-i\sin (n\theta )]$$ $$=2\cos (n\theta )$$

Final Answer: $$\boxed{2\cos (n\theta )}$$

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b) In a group of 12 students, 8 are boys and remaining girls. In how many ways can 5 students be selected for a quiz competition so as to include at most three girls?

Solution:

Given:

• Total students = 12
• Number of boys = 8
• Number of girls = 12 - 8 = 4

We need to select 5 students such that the number of girls is at most three. This means the number of girls in the selection can be 0, 1, 2, or 3.

We calculate the number of ways for each case and sum them up.

Case 1: 0 girls and 5 boys

• Number of ways to choose 0 girls from 4: $\left(\genfrac{}{}{0ex}{}{4}{0}\right)=1$
• Number of ways to choose 5 boys from 8: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=56$
• Total ways for this case: $1\times 56=56$

Case 2: 1 girl and 4 boys

• Number of ways to choose 1 girl from 4: $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$
• Number of ways to choose 4 boys from 8: $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$
• Total ways for this case: $4\times 70=280$

Case 3: 2 girls and 3 boys

• Number of ways to choose 2 girls from 4: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$
• Number of ways to choose 3 boys from 8: $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$
• Total ways for this case: $6\times 56=336$

Case 4: 3 girls and 2 boys

• Number of ways to choose 3 girls from 4: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$
• Number of ways to choose 2 boys from 8: $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$
• Total ways for this case: $4\times 28=112$

Total number of ways:

$$56+280+336+112=784$$

Final Answer: $$\boxed{784}$$

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c) Prove by the method of principle of mathematical induction that $n^3+2n$ is divisible by 3.

Solution:

We will prove the statement using the Principle of Mathematical Induction.

Statement to prove: For all positive integers $n$, $n^3+2n$ is divisible by 3.

Base Case (n = 1):

$$1^3+2(1)=1+2=3$$ 3 is divisible by 3. Thus, the statement holds for $n=1$.

Inductive Step:

Assume that the statement is true for some positive integer $k$, i.e., $$k^3+2k\text{ is divisible by 3.}$$ We need to show that the statement is true for $n=k+1$, i.e., $$(k+1{)}^3+2(k+1)\text{ is divisible by 3.}$$

Let's expand $(k+1{)}^3+2(k+1)$: $$(k+1{)}^3+2(k+1)=k^3+3k^2+3k+1+2k+2$$ $$=k^3+3k^2+5k+3$$

Now, group the terms: $$=(k^3+2k)+(3k^2+3k+3)$$

By the inductive hypothesis, $k^3+2k$ is divisible by 3. Also, $3k^2+3k+3=3(k^2+k+1)$, which is clearly divisible by 3. Therefore, the entire expression is divisible by 3.

Thus, the statement holds for $n=k+1$.

Conclusion:

By the Principle of Mathematical Induction, $n^3+2n$ is divisible by 3 for all positive integers $n$.

Final Answer: The statement is proven by induction.