Binomial Theorem 2 Question 17
20.
Prove that $\left({ }^{2 n} C_{0}\right)^{2}-\left({ }^{2 n} C_{1}\right)^{2}+\left({ }^{2 n} C_{2}\right)^{2}-\ldots+\left({ }^{2 n} C_{2 n}\right)^{2}$ $=(-1)^{n} \cdot{ }^{2 n} C_{n}$.
$(1978,4$ M)
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Solution:
- $(1+x)^{2 n} (1-\frac{1}{x})^{2 n}$
$ \begin{aligned} = & {\left[{ }^{2 n} C_{0}+\left({ }^{2 n} C_{1}\right) x+\left({ }^{2 n} C_{2}\right) x^{2}+\ldots+\left({ }^{2 n} C_{2 n}\right) x^{2 n}\right] } \\ & \times[{ }^{2 n} C_{0}-\left({ }^{2 n} C_{1}\right) \frac{1}{x}+\left({ }^{2 n} C_{2}\right) \frac{1}{x^{2}}+\ldots+\left({ }^{n} C_{2 n}\right) \frac{1}{x^{2 n}}] \end{aligned} $
Independent terms of $x$ on RHS
$ \begin{aligned} & =\left({ }^{2 n} C_{0}\right)^{2}-\left({ }^{2 n} C_{1}\right)^{2}+\left({ }^{2 n} C_{2}\right)^{2}-\ldots+\left({ }^{2 n} C_{2 n}\right)^{2} \\ \text { LHS } & =(1+x)^{2 n} (\frac{x-1}{x}){ }^{2 n}=\frac{1}{x^{2 n}}\left(1-x^{2}\right)^{2 n} \end{aligned} $
Independent term of $x$ on the LHS $=(-1)^{n} \cdot{ }^{2 n} C_{n}$.
