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  4. If n is an even integer, then the value of nC0+nC2+nC4+....+nCn equals to

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Q. If $n$ is an even integer, then the value of $ ^{n}{{C}_{0}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{4}}+....{{+}^{n}}{{C}_{n}} $ equals to

J & K CETJ & K CET 2012Binomial Theorem

Solution:

We know that,
$ {{(1+x)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+....{{+}^{n}}{{C}_{n}}{{x}^{n}} $ ..(i)
and $ {{(1-x)}^{n}}{{=}^{n}}{{C}_{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}} $
$ +...+{{(-1)}^{n}}{{C}_{n}}{{x}^{n}} $ ..(ii)
On adding Eqs. (i) and (ii), we get
$ {{(1+x)}^{n}}+{{(1-x)}^{n}}=2{{[}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{2}}{{x}^{2}}+....{{+}^{n}}{{C}_{n}}{{x}^{n}}] $
$ ^{n}{{C}_{0}}{{+}^{n}}{{C}_{2}}+....{{+}^{n}}{{C}_{n}}=\frac{{{2}^{n}}}{2} $
$ ={{2}^{n-1}} $