Binomial Theorem Ques 30
Prove that $\sum_{r=1}^{k}(-3)^{r-1}{ }^{3}{ }^{n} C_{2 r-1}=0$, where $k=(3 n) / 2$ and $n$ is an even positive integer.
(1993, 5M)
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Solution:
Formula:
Important results on binomial coefficients :
- Since, $n$ is an even positive integer, we can write
$ n=2 m, m=1,2,3, \ldots $
Also, $\quad k=\frac{3 n}{2}=\frac{3(2 m)}{2}=3 m \quad \quad\quad\therefore \quad S=\sum_{r=1}^{3 m}(-3)^{r-1} \cdot{ }^{6 m} C_{2 r-1}$
i.e. $\quad S=(-3)^{0} \quad { }^{6 m} C_{1}+(-3) \quad { }^{6 m} C_{3}+\ldots$ $ +(-3)^{3 m-1} \cdot{ }^{6 m} C_{3 m-1} \quad …….(i) $
From the binomial expansion, we write
$ \begin{aligned} (1+x)^{6 m} & ={ }^{6 m} C_{0}+{ }^{6 m} C_{1} x+{ }^{6 m} C_{2} x^{2}+\ldots \\ & { }^{6 m} C_{6 m-1} x^{6 m-1}+{ }^{6 m} C_{6 m} x^{6 m} \quad …….(ii)\\ (1-x)^{6 m}= & { }^{6 m} C_{0}+{ }^{6 m} C_{1}(-x)+{ }^{6 m} C_{2}(-x)^{2}+\ldots \\ & +{ }^{6 m} C_{6 m-1}(-x)^{6 m-1}+{ }^{6 m} C_{6 m}(-x)^{6 m} \quad …….(iii) \end{aligned} $
On subtracting Eq. (iii) from Eq. (ii), we get
$(1+x)^{6 m}-(1-x)^{6 m}=2\left[{ }^{6 m} C_{1} x+{ }^{6 m} C_{3} x^{3}\right. $ $\left.+{ }^{6 m} C_{5} x^{5}+\ldots+{ }^{6 m} C_{6 m-1} x^{6 m-1}\right] $
$\Rightarrow \frac{(1+x)^{6 m}-(1-x)^{6 m}}{2 x}={ }^{6 m} C_{1}+{ }^{6 m} C_{3} x^{2}+{ }^{6 m} C_{5} x^{4}+\ldots $ $+{ }^{6 m} C_{6 m-1} x^{6 m-2}$
$ \text { Let } \quad x^{2}=y $
$\Rightarrow \quad \frac{(1+\sqrt{y})^{6 m}-(1-\sqrt{y})^{6 m}}{2 \sqrt{y}}={ }^{6 m} C_{1}+{ }^{6 m} C_{3} y$
$ +{ }^{6 m} C_{5} y^{2}+\ldots+{ }^{6 m} C_{6 m-1} y^{3 m-1} $
For the required sum we have to put $y=-3$ in RHS.
$ \begin{aligned} \therefore \quad S & =\frac{(1+\sqrt{-3})^{6 m}-(1-\sqrt{-3})^{6 m}}{2 \sqrt{-3}} \\ & =\frac{(1+i \sqrt{3})^{6 m}-(1-i \sqrt{3})^{6 m}}{2 i \sqrt{3}} \quad …….(iv) \end{aligned} $
Let $\quad z=1+i \sqrt{3}=r(\cos \theta+i \sin \theta)$
$\Rightarrow \quad r=|z|=\sqrt{1+3}=2$
and $\quad \theta=\pi / 3$
Now, $\quad z^{6 m}=[r(\cos \theta+i \sin \theta)]^{6 m}$
$ =r^{6 m}(\cos 6 m \theta+i \sin 6 m \theta) $
Again, $\quad \bar{z}=r(\cos \theta-i \sin \theta)$
and $\quad(\bar{z})^{6 m}=r^{6 m}(\cos 6 m \theta-i \sin 6 m \theta)$
$\Rightarrow \quad z^{6 m}-\bar{z}^{6 m}=r^{6 m}(2 i \sin 6 m \theta) \quad …….(v)$
From Eq. (i),
$ \begin{aligned} S & =\frac{z^{6 m}-\bar{z}^{6 m}}{2 i \sqrt{3}}=\frac{r^{6 m}(2 i \sin 6 m \theta)}{2 i \sqrt{3}} \\ & =\frac{2^{6 m} \sin 6 m \theta}{\sqrt{3}} \\ & =0 \text { as } m \in z, \text { and } \theta=\pi / 3 \end{aligned} $
BETA
