# q help (1 Viewer)

#### Anounymouse352

##### New Member
Hey, guys any help with following questions would be greatly appreciated

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following working help with the question in the fourth attachment?

#### Lith_30

##### Active Member
For the second attachment

part i:

$\bg_white \\LHS-RHS=\frac{1}{a}+\frac{1}{b}-\frac{2}{3}\\\\=\frac{3b+3a-2ab}{3ab}\\\\=\frac{3(b+a)-2ab}{3ab}\\\\=\frac{0.5(b+a)(b+a)-2ab}{3ab}\\\\=\frac{\frac{(b^2+2ab+a^2)}{2}-2ab}{3ab}\\\\=\frac{\frac{(b^2+a^2)}{2}-ab}{3ab}\\\\\geq\frac{\sqrt{a^2b^2}-ab}{3ab}\ \text{AM GM inequality}\\\\\geq0$

hence $\bg_white \frac{1}{a}+\frac{1}{b}\geq\frac{2}{3}$

For part ii it is pretty much the same process, do LHS-RHS, combine all the fractions then use AM GM to make it equal zero.

#### ExtremelyBoredUser

##### hiiii
Hey, guys any help with following questions would be greatly appreciated

View attachment 35323
i)
$\bg_white (1 + i)^n = (\sqrt{2})^n \times e^{i(\frac{n\pi}{4})}$
$\bg_white (1 - i)^n = (\sqrt{2})^n \times e^{-i(\frac{n\pi}{4})}$

Summing these together, we get the RHS;

let theta equal npi/4
$\bg_white RHS = (\sqrt{2})^n \times (e^{i\theta} + e^{-i\theta})$
$\bg_white RHS = (\sqrt{2})^n \times (e^{i\theta} + e^{-i\theta})$
$\bg_white RHS = (\sqrt{2})^n \times (\cos(\theta) + i\sin(\theta) + \cos(-\theta) + i\sin(-\theta)$
$\bg_white RHS = (\sqrt{2})^n \times (\cos(\theta) + i\sin(\theta) + \cos(\theta) - i\sin(\theta)$
$\bg_white RHS = (\sqrt{2})^n \times (2\cos(\theta)$
$\bg_white RHS = (\sqrt{2})^n \times ((\sqrt{2})^n \times \cos(\theta)$
$\bg_white RHS = (\sqrt{2})^{n+2} \times \cos(\frac{n\pi}{4})$

#### Lith_30

##### Active Member
for part ii)

$\bg_white \\(1+i)^n+(1-i)^n\ \text{Where n is a multiple of 4}\\\\={n\choose0}+{n\choose1}i-{n\choose2}-{n\choose3}i+{n\choose4}-...+{n\choose{n}}+{n\choose0}+{n\choose1}(-1)i-{n\choose2}-{n\choose3}(-1)^3i+{n\choose4}+...+{n\choose{n}}\ \text{Binomial expansion}\\\\={n\choose0}+{n\choose1}i-{n\choose2}-{n\choose3}i+{n\choose4}-...+{n\choose{n}}+{n\choose0}-{n\choose1}i-{n\choose2}+{n\choose3}i+{n\choose4}+...+{n\choose{n}}$

As you may be able to see all the imaginary components cancel each other out and the real components add onto each other, as it should.

$\bg_white \\2\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=(1+i)^n+(1-i)^n\\\\2\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=(\sqrt{2})^{n+2}\cos(\frac{n\pi}{4})\\\\\left({n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}\right)=\sqrt{2}^n\cos(\frac{n\pi}{4})$

we have the $\bg_white \sqrt{2}^n$, but what can we do with the cos function to get what we want??

Since we know that n is a multiple of 4, we can let $\bg_white n=4k, k\in\mathbb{Z},k\geq0$

hence $\bg_white \cos(\frac{n\pi}{4})=\cos(k\pi)=1,\ -1,\ 1,\ -1...$ (starting at n=0)

therefore we can say that

$\bg_white \\\cos(k\pi)=(-1)^k\\\\\cos(\frac{n\pi}{4})=(-1)^{\frac{n}{4}}$

therefore $\bg_white {n\choose0}-{n\choose2}+{n\choose4}-...+{n\choose{n}}=\sqrt{2}^n(-1)^{\frac{n}{4}}$ as required