Answer:
Hello! I know I am late but here is the answer for anyone else who might need it.
D. 1/2-i√3/2
Step-by-step explanation:
I got it right on edge. Hope this helps!
The value of given expression is [tex]\frac{1}{2}-\frac{\sqrt{3} i}{2}[/tex].
What is binomial theorem?
The expression of the form (a+b)^n is expand by binomial theorem.
The expansion is [tex](a+b)^{n}=\sum_{r=0}^{n}{ }^{n} C_{r} a^{n-r} b^{r}[/tex].
Use a=1/2 and b=[tex]b=\frac{i\sqrt{3} }{2}[/tex] into the expansion of binomial.
[tex]\left(\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)^{5}=\sum_{k=0}^{5}\left(\begin{array}{l}5 \\k\end{array}\right)\left(\frac{1}{2}\right)^{5-k}\left(\frac{\sqrt{3} i}{2}\right)^{k}[/tex]
When k=0 then, [tex]\left(\begin{array}{l}5 \\0\end{array}\right)\left(\frac{1}{2}\right)^{5-0}\left(\frac{\sqrt{3} i}{2}\right)^{0}=\frac{5 !}{(5-0) ! 0 !}\left(\frac{1}{2}\right)^{5-0}\left(\frac{\sqrt{3} i}{2}\right)^{0}=\frac{1}{32}[/tex]
When k=1 then, [tex]\left(\begin{array}{l}5 \\1\end{array}\right)\left(\frac{1}{2}\right)^{5-1}\left(\frac{\sqrt{3} i}{2}\right)^{1}=\frac{5 !}{(5-1) ! 1 !}\left(\frac{1}{2}\right)^{5-1}\left(\frac{\sqrt{3} i}{2}\right)^{1}=\frac{5 \sqrt{3} i}{32}[/tex]
When k=2 then, [tex]\left(\begin{array}{l}5 \\2\end{array}\right)\left(\frac{1}{2}\right)^{5-2}\left(\frac{\sqrt{3} i}{2}\right)^{2}=\frac{5 !}{(5-2) ! 2 !}\left(\frac{1}{2}\right)^{5-2}\left(\frac{\sqrt{3} i}{2}\right)^{2}=-\frac{15}{16}[/tex]
When k=3 then, [tex]\left(\begin{array}{l}5 \\3\end{array}\right)\left(\frac{1}{2}\right)^{5-3}\left(\frac{\sqrt{3} i}{2}\right)^{3}=\frac{5 !}{(5-3) ! 3 !}\left(\frac{1}{2}\right)^{5-3}\left(\frac{\sqrt{3} i}{2}\right)^{3}=-\frac{15 \sqrt{3} i}{16}[/tex]
When k=4 then, [tex]\left(\begin{array}{l}5 \\4\end{array}\right)\left(\frac{1}{2}\right)^{5-4}\left(\frac{\sqrt{3} i}{2}\right)^{4}=\frac{5 !}{(5-4) ! 4 !}\left(\frac{1}{2}\right)^{5-4}\left(\frac{\sqrt{3} i}{2}\right)^{4}=\frac{45}{32}[/tex]
When k=5 then, [tex]\left(\begin{array}{l}5 \\5\end{array}\right)\left(\frac{1}{2}\right)^{5-5}\left(\frac{\sqrt{3} i}{2}\right)^{5}=\frac{5 !}{(5-5) ! 5 !}\left(\frac{1}{2}\right)^{5-5}\left(\frac{\sqrt{3} i}{2}\right)^{5}=\frac{9 \sqrt{3} i}{32}[/tex],
Now we will add all the results to get the expansion of given expression,
[tex]\left(\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)^{5}=\frac{1}{32}+\frac{5 \sqrt{3} i}{32}-\frac{15}{16}-\frac{15 \sqrt{3} i}{16}+\frac{45}{32}+\frac{9 \sqrt{3} i}{32}[/tex]
On simplifying,
[tex]\left(\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)^{5}=\frac{1}{2}-\frac{\sqrt{3} i}{2}[/tex]
Therefore, the value of given expression is [tex]\frac{1}{2}-\frac{\sqrt{3} i}{2}[/tex].
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