Question

If$i=\sqrt{-1}$. If $\left[\frac{{\left(-1+i\sqrt{3}\right)}^{21}}{{(1–i)}^{24}}\right]+\left[\frac{{\left(1+i\sqrt{3}\right)}^{21}}{{(1+i)}^{24}}\right]=k$, and $n=\left[\left|k\right|\right]$ be the greatest integral part of $\left|k\right|$. Then

$\begin{array}{l}su{m}_{j=0}^{n+5}(j+5{)}^2–su{m}_{j=0}^{n+5}(j+5)\end{array}$ equals to__

Answer 1

Explanation for the correct option:

Finding the value of $\begin{array}{l}su{m}_{j=0}^{n+5}(j+5{)}^2–su{m}_{j=0}^{n+5}(j+5)\end{array}$

We know the polar form of $z=x+iy$

$\Rightarrow r\left(\cos \theta +i\sin \theta \right)=r{e}^{i\theta }\text{and}\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ where, $r=\sqrt{x^2+y^2}$

Since

$$\begin{gathered} r=\sqrt{1^2+{\left(\sqrt{3}\right)}^2} \\ =\sqrt{1+3} \\ =\sqrt{4} \\ =2 \end{gathered}$$

and

$$\begin{gathered} \theta ={\tan }^{-1}\left(-\sqrt{3}\right) \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Then,

$-1+i\sqrt{3}=2e^{i\left(\frac{2\mathrm{\pi }}{3}\right)}......\left(1\right)$

Similarly,

$1-i=\sqrt{2}{e}^{-i\left(\frac{\mathrm{\pi }}{4}\right)}......\left(2\right)$

And

$1+i=\sqrt{2}{e}^{i\left(\frac{\mathrm{\pi }}{4}\right)}......\left(3\right)$

Now consider the given Equation as,

$\left[\frac{{\left(-1+i\sqrt{3}\right)}^{21}}{{(1–i)}^{24}}\right]+\left[\frac{{\left(1+i\sqrt{3}\right)}^{21}}{{(1+i)}^{24}}\right]=k$

Substitute the value of Equation (1), (2) and (3) in the above Equation becomes,

$$\begin{gathered} k=\frac{{\left[2e^{i\left(\frac{2\mathrm{\pi }}{3}\right)}\right]}^{21}}{{\left[\sqrt{2}{e}^{-i\left(\frac{\mathrm{\pi }}{4}\right)}\right]}^{24}}+\frac{{\left[2e^{i\left(\frac{\mathrm{\pi }}{3}\right)}\right]}^{21}}{{\left[\sqrt{2}{e}^{i\left(\frac{\mathrm{\pi }}{4}\right)}\right]}^{24}} \\ k=\frac{\left[2^{21}{e}^{i\left(\frac{2\mathrm{\pi }}{3}\times 21\right)}\right]}{\left[{\left(\sqrt{2}\right)}^{24}{e}^{-i\left({\displaystyle \frac{\mathrm{\pi }}{4}}\times 24\right)}\right]}+\frac{\left[2^{21}{e}^{i\left(\frac{\mathrm{\pi }}{3}\times 21\right)}\right]}{\left[{\left(\sqrt{2}\right)}^{24}{e}^{i\left({\displaystyle \frac{\mathrm{\pi }}{4}}\times 24\right)}\right]} \\ k=\frac{\left[2^{21}{e}^{i\left(2\mathrm{\pi }\times 7\right)}\right]}{\left[{\left(2\right)}^{{\displaystyle \frac{1}{2}}\times 24}{e}^{-i\left({\displaystyle 6\mathrm{\pi }}\right)}\right]}+\frac{\left[2^{21}{e}^{i\left(2\mathrm{\pi }\times 7\right)}\right]}{\left[{\left(2\right)}^{{\displaystyle \frac{1}{2}}\times 24}{e}^{i\left(6\mathrm{\pi }\right)}\right]} \\ k=\frac{2^{21}{e}^{14\mathrm{\pi i}}}{2^{12}{e}^{-6\mathrm{\pi i}}}+\frac{2^{21}{e}^{7\mathrm{\pi i}}}{2^{12}{e}^{6\mathrm{\pi i}}} \\ k=2^{21-12}{e}^{14\mathrm{\pi i}+6\mathrm{\pi i}}+2^{21-12}{e}^{7\mathrm{\pi i}-6\mathrm{\pi i}} \\ k=2^9{e}^{20\mathrm{\pi i}}+2^9{e}^{\mathrm{\pi i}} \\ k=2^9\left(1\right)+2^9\left(-1\right) \\ k=2^9-2^9 \\ k=0 \end{gathered}$$

Since $n=\left[\left|k\right|\right]$

Therefore $n=0$

So now,

$$\begin{gathered} \Rightarrow \sum _{j=0}^5{(j+5)}^2–\sum _{j=0}^5(j+5)=\left[5^2+6^2+7^2+8^2+9^2+{10}^2\right]–[5+6+7+8+9+10] \\ =\left[\right(1^2+2^2+\dots +{10}^2)–(1^2+2^2+3^2+4^2)]–\left[\right(1+2+3+\dots +10)–(1+2+3+4)] \\ =(385–30)-[55–10] \\ =355–45 \\ =310 \end{gathered}$$

Therefore, $\begin{array}{l}{\mathbf{sum}}_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}\mathbf{+}\mathbf{5}}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{5}{\mathbf{)}}^{\mathbf{2}}\mathbf{–}{\mathbf{sum}}_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}\mathbf{+}\mathbf{5}}\mathbf{(}\mathbf{j}\mathbf{+}\mathbf{5}\mathbf{)}\end{array}\mathbf{=}\mathbf{310}$