Question

Prove that $ \sum\limits_{r=1}^{k}{{{\left( -3 \right)}^{r}}{}^{3n}{{C}_{2r-1}}}=0 $ where $ k=\dfrac{3n}{2} $ and $ n $ is an even integer.\[\]

Answer 1

Hint: We take $ n=2m $ and expand the give summation $ S=\sum\limits_{r=1}^{k}{{{\left( -3 \right)}^{r}}{}^{3n}{{C}_{2r-1}}} $ at left hand side. We subtract corresponding sides of $ {{\left( 1-x \right)}^{6m}} $ from $ {{\left( 1-x \right)}^{6m}} $ , put $ {{x}^{2}}=y $ in the subtracted equation and the put $ y=-3 $ to get summation $ S $ . We find at the left hand side two complex numbers $ \left( 1+i\sqrt{3} \right),\left( 1-i\sqrt{3} \right) $ which we convert $ z=x+iy $ to $ \left( r,\theta \right) $ from using $ z=r\cos \theta +ir\sin \theta $ where $ r=\sqrt{{{x}^{2}}+{{y}^{2}}},\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right) $ . We add both the complex numbers in $ \left( r,\theta \right) $ from to get $ S=0 $ . \[\]

Complete step by step answer:
We know that we can use the binomial theorem (or binomial expansion) to describe the algebraic expansion of the power of binomials. If $ x,y $ are the two terms of binomial with some positive integral power $ n $ then the binomial expansion is given by;
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{0}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{0}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}\]
The above expression is called binomial formula or binomial identity. We know that we can convert any complex number $ z=x+iy $ into $ \left( r,\theta \right) $ form as ;\[z=r\cos \theta +ir\sin \theta \text{ where }r=\sqrt{{{x}^{2}}+{{y}^{2}}},\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
The complex number $ \overline{z}=z-iy $ is called conjugate of $ z=x+iy $ whose $ \left( r,\theta \right) $ form is;
\[z=r\cos \theta -ir\sin \theta \text{ }\]
We are given in the question to prove
\[\sum\limits_{r=1}^{k}{{{\left( -3 \right)}^{r}}{}^{3n}{{C}_{2r-1}}}=0\]
Let us denote the summation on the left hand side as $ S $ . We have;
\[S=\sum\limits_{r=1}^{k}{{{\left( -3 \right)}^{r}}{}^{3n}{{C}_{2r-1}}}\]
We put $ n=2m $ for some integer $ m $ since we are given $ n $ is even. Then we have $ k=\dfrac{3\times 2m}{2}=3m $ . We put $ n=3m,k=2m $ in the above step and expand the summation to have;
\[\begin{align}
  & \Rightarrow S=\sum\limits_{r=1}^{2m}{{{\left( -3 \right)}^{r}}{}^{6m}{{C}_{2r-1}}} \\
 & \Rightarrow S={{\left( -3 \right)}^{0}}{}^{6m}{{C}_{0}}+\left( -3 \right){}^{6m}{{C}_{3}}+{{\left( -3 \right)}^{2}}{}^{6m}{{C}_{5}}+...+{{\left( -3 \right)}^{3m-1}}{}^{6m}{{C}_{3m-1}}.....\left( 1 \right) \\
\end{align}\]
We put $ x=1,y=x,n=6m $ in the binomial expansion to have.
\[{{\left( 1+x \right)}^{6m}}={}^{6m}{{C}_{0}}+{}^{6m}{{C}_{1}}x+{}^{6m}{{C}_{2}}{{x}^{2}}+...{}^{6m}{{C}_{6m-1}}{{x}^{6m-1}}+{}^{6m}{{C}_{6m}}{{x}^{6m}}.........\left( 2 \right)\]
We put $ x=1,y=-x,n=6m $ in the binomial expansion to have;
\[{{\left( 1-x \right)}^{6m}}={}^{6m}{{C}_{0}}+{}^{6m}{{C}_{1}}\left( -x \right)+{}^{6m}{{C}_{2}}{{\left( -x \right)}^{2}}+...{}^{6m-1}{{C}_{6m-1}}{{\left( -x \right)}^{6m-1}}+{}^{6m}{{C}_{6m}}{{\left( -x \right)}^{6m}}.........\left( 3 \right)\]
We subtract equation (3) from equation (2) and see that all the terms with even power of $ x $ will vanish. We have;
\[\begin{align}
  & {{\left( 1+x \right)}^{6m}}-{{\left( 1-x \right)}^{6m}}={}^{6}{{C}_{1}}\left( x-\left( -x \right) \right)+{}^{6}{{C}_{3}}\left( {{x}^{3}}-{{\left( -x \right)}^{3}} \right)+...+{}^{6m}{{C}_{6m-1}}\left( {{x}^{6m-1}}-{{\left( -x \right)}^{6m-1}} \right) \\
 & \Rightarrow {{\left( 1+x \right)}^{6m}}-{{\left( 1-x \right)}^{6m}}={}^{6}{{C}_{1}}\cdot 2x+2{{x}^{3}}{}^{6}{{C}_{3}}\cdot 2{{x}^{3}}+...+{}^{6m}{{C}_{6m-1}}\cdot 2{{x}^{6m-1}} \\
\end{align}\]
We divide both sides of the above equation by $ 2x $ to have;
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( 1+x \right)}^{6m}}-{{\left( 1-x \right)}^{6m}}}{2x}={}^{6}{{C}_{1}}+{}^{6}{{C}_{3}}{{x}^{2}}+...+{}^{6m}{{C}_{6m-1}}{{x}^{6m-2}} \\
 & \Rightarrow \dfrac{{{\left( 1+x \right)}^{6m}}-{{\left( 1-x \right)}^{6m}}}{2x}={}^{6}{{C}_{1}}+{}^{6}{{C}_{3}}{{x}^{2}}+...+{}^{6m}{{C}_{6m-1}}{{\left( {{x}^{2}} \right)}^{3m-1}} \\
\end{align}\]
We put $ {{x}^{2}}=y $ in the above step to have;
\[\Rightarrow \dfrac{{{\left( 1+\sqrt{y} \right)}^{6m}}-{{\left( 1-\sqrt{y} \right)}^{6m}}}{2\sqrt{y}}={}^{6}{{C}_{1}}+{}^{6}{{C}_{3}}\cdot y+...+{}^{6m}{{C}_{6m-1}}{{\left( y \right)}^{3m-1}}......\left( 4 \right)\]
We observe the right hand above the equation and compare it with the right hand side of equation (1). We see that if we take $ y=-3 $ then the right hand sides of equation (1) and (4) will be the same. So we have
\[\Rightarrow \dfrac{{{\left( 1+\sqrt{-3} \right)}^{6m}}-{{\left( 1-\sqrt{-3} \right)}^{6m}}}{2\sqrt{-3}}={}^{6}{{C}_{1}}+{}^{6}{{C}_{3}}\left( -3 \right)+...+{}^{6m}{{C}_{6m-1}}{{\left( -3 \right)}^{3m-1}}\]
We now choose to equate left hand side of above equation and equation (1) to have;
\[\begin{align}
  & \Rightarrow S=\dfrac{{{\left( 1+\sqrt{-3} \right)}^{6m}}-{{\left( 1-\sqrt{-3} \right)}^{6m}}}{2\sqrt{-3}} \\
 & \Rightarrow S=\dfrac{{{\left( 1+i\sqrt{3} \right)}^{6m}}-{{\left( 1-i\sqrt{3} \right)}^{6m}}}{2i\sqrt{3}}.......\left( 5 \right) \\
\end{align}\]
We see that in the numerator there are two complex numbers which are conjugate. We take $ z=1+i\sqrt{3} $ and convert it into $ \left( r,\theta \right) $ form to have;
\[ \begin{align}
  & r=\sqrt{{{1}^{^{2}}}+{{\left( \sqrt{3} \right)}^{2}}}=\sqrt{4}=2,\theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{3}}{1} \right)=\dfrac{\pi }{3} \\
 & \therefore z=1+i\sqrt{3}=2\cos \left( \dfrac{\pi }{3} \right)+2i\sin \left( \dfrac{\pi }{3} \right) \\
\end{align}\]
We find the conjugate of $ z=1+i\sqrt{3} $ as $ \overline{z}=1-i\sqrt{3} $ and express it $ \left( r,\theta \right) $ form to have;
\[ \overline{z}=1-i\sqrt{3}=2\cos \left( \dfrac{\pi }{3} \right)-2i\sin \left( \dfrac{\pi }{3} \right)\]
We put the values of $ z,\overline{z} $ in equation (5) to have;
\[\begin{align}
  & S=\dfrac{{{\left( 2\cos \left( \dfrac{\pi }{3} \right)+2i\sin \left( \dfrac{\pi }{3} \right) \right)}^{6m}}-{{\left( 2\cos \left( \dfrac{\pi }{3} \right)-2i\sin \left( \dfrac{\pi }{3} \right) \right)}^{6m}}}{2i\sqrt{3}} \\
 & \Rightarrow S=\dfrac{{{2}^{6m}}{{\left( \cos \left( \dfrac{\pi }{3} \right)+i\sin \left( \dfrac{\pi }{3} \right) \right)}^{6m}}-{{2}^{6m}}{{\left( \cos \left( \dfrac{\pi }{3} \right)-i\sin \left( \dfrac{\pi }{3} \right) \right)}^{6m}}}{2i\sqrt{3}} \\
\end{align}\]
We use the identity $ {{\left( \cos \theta +i\sin \theta \right)}^{k}}=\cos k\theta +i\sin k\theta $ for $ \theta =\dfrac{\pi }{3},k=6m $ to have;
\[\begin{align}
  & \Rightarrow S=\dfrac{{{2}^{6m}}\left( \cos 2\pi +i\sin 2\pi \right)-{{2}^{6m}}\left( \cos 2\pi -i\sin 2\pi \right)}{2i\sqrt{3}} \\
 & \Rightarrow S=\dfrac{{{2}^{6m}}\left( \cos 2\pi +i\sin 2\pi -\cos 2\pi +i\sin 2\pi \right)}{2i\sqrt{3}} \\
 & \Rightarrow S=\dfrac{{{2}^{6m}}\cdot 2i\sin 2\pi }{2i\sqrt{3}}=0\left( \because \sin 2\pi =0 \right) \\
 & \Rightarrow \sum\limits_{r=1}^{k}{{{\left( -3 \right)}^{r}}{}^{3n}{{C}_{2r-1}}}=0 \\
\end{align}\]
Hence it is proved. \[\].

Note:
We note that the form $ \left( r,\theta \right) $ is called polar form where $ r $ is the modulus of the complex number and $ \theta $ is the argument of the complex number $ z=x+iy $ where $ x $ is called the real part and $ y $ is called the imaginary part. We can alternatively solve by putting $ x=i\sqrt{3} $ in the expansion of $ {{\left( 1+x \right)}^{6m}} $ and then comparing the imaginary parts of both sides of the equation.