Question

Find $K, m$ for which $\left( {{}^m{C_0} + {}^m{C_1} - {}^m{C_2} - {}^m{C_3}} \right) + \left( {{}^m{C_4} + {}^m{C_5} - {}^m{C_6} - {}^m{C_7}} \right) + ... = 0$ is true if and only if for some positive integer $k,m = ?$
A) $4k$
B) $4k + 1$
C) $4k - 1$
D) $4k + 2$

Answer 1

Here, we will first consider the binomial expansion for the polar coordinates. Then we will add them and subtract them to get two equations. We will then use the trigonometric identities and rules of exponents to find the equation which is of the given form of equation. By equating the trigonometric equations, we will find the variable with respect to the integer.

Formula Used:
We will use the following formula:

1. $\cos m\theta + \sin m\theta = \sqrt 2 \sin \left( {m\theta + \dfrac{\pi }{4}} \right)$
2. Exponent product rule: ${a^m} \cdot {a^n} = {a^{m + n}}$
3. Trigonometric Ratio: $\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$

Complete step by step solution:
Let us consider the binomial expansion for the polar coordinates and thus we get
${\left( {\cos \theta - i\sin \theta } \right)^m} = {}^m{C_0}{\cos ^m}\theta - {}^m{C_1}{\cos ^{m - 1}}\theta \left( {i\sin \theta } \right) + .............. + {}^m{C_m}{\left( { - i\sin \theta } \right)^m}$ ………………..$\left( 1 \right)$
${\left( {\cos \theta + i\sin \theta } \right)^m} = {}^m{C_0}{\cos ^m}\theta + {}^m{C_1}{\cos ^{m - 1}}\theta \left( {i\sin \theta } \right) + .............. + {}^m{C_m}{\left( {i\sin \theta } \right)^m}$ …………… $\left( 2 \right)$
Now, by adding equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we get
Since the even terms of the binomial expansion are opposite in signs, so that they cancel each other.
$2\cos m\theta = \left( {2{}^m{C_0}{{\cos }^m}\theta - 2{}^m{C_2}{{\cos }^{m - 2}}\theta {{\sin }^2}\theta + ......} \right)$
Factoring out common terms, we get
$\Rightarrow 2\cos m\theta = 2\left( {{}^m{C_0}{{\cos }^m}\theta - {}^m{C_2}{{\cos }^{m - 2}}\theta {{\sin }^2}\theta + ......} \right)$
$\Rightarrow \cos m\theta = \left( {{}^m{C_0}{{\cos }^m}\theta - {}^m{C_2}{{\cos }^{m - 2}}\theta {{\sin }^2}\theta + ......} \right)$ …………………………………………$\left( 3 \right)$
Now, by subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$, we get
$2i\sin m\theta = \left( {2{}^m{C_1}{{\cos }^{m - 1}}\theta i\sin \theta - 2{}^m{C_3}{{\cos }^{m - 3}}\theta i{{\sin }^3}\theta + ......} \right)$
Since the odd terms of the binomial expansion are opposite in signs, so that they cancel each other.
$\Rightarrow 2i\sin m\theta = 2i\left( {{}^m{C_1}{{\cos }^{m - 1}}\theta \sin \theta - {}^m{C_3}{{\cos }^{m - 3}}\theta {{\sin }^3}\theta + ......} \right)$
By taking the common factor, we get
$\Rightarrow \sin m\theta = \left( {{}^m{C_1}{{\cos }^{m - 1}}\theta \sin \theta - {}^m{C_3}{{\cos }^{m - 3}}\theta {{\sin }^3}\theta + ......} \right)$ ……………………………………………………… $\left( 4 \right)$
Now, by adding equation $\left( 3 \right)$ and equation $\left( 4 \right)$, we get
$\cos m\theta + \sin m\theta = \left[ {{}^m{C_0}{{\cos }^m}\theta + {}^m{C_1}{{\cos }^{m - 1}}\theta \sin \theta - {}^m{C_2}{{\cos }^{m - 2}}\theta {{\sin }^2}\theta - {}^m{C_3}{{\cos }^{m - 3}}\theta {{\sin }^3}\theta ..........} \right]$
We know that $\cos m\theta + \sin m\theta = \sqrt 2 \sin \left( {m\theta + \dfrac{\pi }{4}} \right)$.
By substituting $\cos m\theta + \sin m\theta = \sqrt 2 \sin \left( {m\theta + \dfrac{\pi }{4}} \right)$, we get
$\Rightarrow \sqrt 2 \sin \left( {m\theta + \dfrac{\pi }{4}} \right) = \left[ {{}^m{C_0}{{\cos }^m}\theta + {}^m{C_1}{{\cos }^{m - 1}}\theta \sin \theta - {}^m{C_2}{{\cos }^{m - 2}}\theta {{\sin }^2}\theta - {}^m{C_3}{{\cos }^{m - 3}}\theta {{\sin }^3}\theta ..........} \right]$
By substituting $\theta = \dfrac{\pi }{4}$ in the above equation, we get
\Rightarrow \sqrt 2 \sin \left( {m\dfrac{\pi }{4} + \dfrac{\pi }{4}} \right) = \left[ {{}^m{C_0}{{\cos }^m}\dfrac{\pi }{4} + {}^m{C_1}{{\cos }^{m - 1}}\dfrac{\pi }{4}\sin \dfrac{\pi }{4} - {}^m{C_2}{{\cos }^{m - 2}}\dfrac{\pi }{4}{{\sin }^2}\dfrac{\pi }{4}..........} \right]$$$$ \Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = \left[ {{}^m{C_0}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m} + {}^m{C_1}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{m - 1}} \cdot {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^1} - {}^m{C_2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{m - 2}} \cdot {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}.........} \right]
By substituting the known values for the trigonometric equation and using the exponent product rule, we get$\Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = \left[ {{}^m{C_0}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m} + {}^m{C_1}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{m - 1 + 1}} - {}^m{C_2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{m - 2 + 2}} - {}^m{C_3}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{m - 3 + 3}}..........} \right]$
By adding and subtracting the terms in the exponents, we get
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = \left[ {{}^m{C_0}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m} + {}^m{C_1}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m} - {}^m{C_2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m} - {}^m{C_3}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^m}..........} \right]$
By taking out the common factors and rewriting the surds in the form of exponents, we get
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = \dfrac{1}{{{2^{\dfrac{m}{2}}}}}\left[ {\left( {{}^m{C_0} + {}^m{C_1} - {}^m{C_2} - {}^m{C_3}} \right)... + \left( {{}^m{C_{m - 3}} + {}^m{C_{m - 2}} - {}^m{C_{m - 1}} - {}^m{C_m}} \right)} \right]$
It is given that $\left( {{}^m{C_0} + {}^m{C_1} - {}^m{C_2} - {}^m{C_3}} \right) + \left( {{}^m{C_4} + {}^m{C_5} - {}^m{C_6} - {}^m{C_7}} \right) + ... = 0$. Therefore,
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = \dfrac{1}{{{2^{\dfrac{m}{2}}}}}\left( 0 \right)$
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{{\left( {m + 1} \right)\pi }}{4}} \right) = 0$
By rewriting the equation, we get
$\Rightarrow \dfrac{{\left( {m + 1} \right)\pi }}{4} = k\pi$
By cancelling the similar terms, we get
$\Rightarrow \dfrac{{m + 1}}{4} = k$
By rewriting the equation, we get
$\Rightarrow m + 1 = 4k$
$\Rightarrow m = 4k - 1$ where $k \in {\bf{N}}$
Therefore, $\left( {{}^m{C_0} + {}^m{C_1} - {}^m{C_2} - {}^m{C_3}} \right) + \left( {{}^m{C_4} + {}^m{C_5} - {}^m{C_6} - {}^m{C_7}} \right) + ... = 0$ if and only if for some positive integer $k,$ the value of $m$ is $4k - 1$ where $k \in {\bf{N}}$.

Thus Option (C) is the correct answer.

Note:
We should know that the binomial coefficient uses the concept of combinations. Binomial expansion has the number of terms greater by 1 than the power of the binomial expansion. The sum of the exponents of a Binomial expansion is always $m$ which is the power of the binomial expansion. We should also know the trigonometric formula and ratios while solving the binomial expansion for the variable $m$. Binomial coefficients are the integers which are coefficients in the Binomial Theorem.