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  4. The value of 40C0+40C1+40C2+...+40C20 is

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Q. The value of $ ^{40}{{C}_{0}}{{+}^{40}}{{C}_{1}}{{+}^{40}}{{C}_{2}}+...{{+}^{40}}{{C}_{20}} $ is

Bihar CECEBihar CECE 2015

Solution:

$ ^{40}{{C}_{0}}{{+}^{40}}{{C}_{1}}{{+}^{40}}{{C}_{2}}+...{{+}^{40}}{{C}_{20}} $ $ =\frac{1}{2}[{{2.}^{40}}{{C}_{0}}+{{2.}^{40}}{{C}_{1}}+{{2.}^{40}}{{C}_{2}}+...+{{2.}^{40}}{{C}_{20}}] $ $ =\frac{1}{2}[{{(}^{40}}{{C}_{0}}{{+}^{40}}{{C}_{40}})+{{(}^{40}}{{C}_{1}}{{+}^{40}}{{C}_{39}})+...+ $ $ {{(}^{40}}{{C}_{19}}{{+}^{40}}{{C}_{21}})+{{2}^{40}}{{C}_{20}}] $ $ =\frac{1}{2}[{{\{}^{40}}{{C}_{0}}{{+}^{40}}{{C}_{1}}{{+}^{40}}{{C}_{2}}+...{{+}^{40}}{{C}_{19}}{{+}^{40}}{{C}_{20}}] $ $ {{+}^{40}}{{C}_{21}}\}{{+}^{40}}{{C}_{20}}] $ $ =\frac{1}{2}\left[ {{2}^{40}}+\frac{40!}{{{(20!)}^{2}}} \right]={{2}^{39}}+\frac{1}{2}\frac{40!}{{{(20!)}^{2}}} $