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Q. The sum of the coefficients in the binomial expansion of $\left(\frac{1}{x}+2x\right)^{6}$ is equal to

KEAMKEAM 2014Binomial Theorem

Solution:

$\left(\frac{1}{x}+2 x\right)^{6}=\left(\frac{1}{x}\right)^{6}+{ }^{6} C_{1}\left(\frac{1}{x}\right)^{5}(2 x)^{1}+{ }^{6} C_{2}\left(\frac{1}{x}\right)^{4}$
$(2 x)^{2}+{ }^{6} C_{3}\left(\frac{1}{x}\right)^{3}(2 x)^{3}+{ }^{6} C_{4}\left(\frac{1}{x}\right)^{2}(2 x)^{4}$
$+{ }^{6} C_{5}\left(\frac{1}{x}\right)(2 x)^{5}+{ }^{6} C_{6}(2 x)^{6}$
$=\left(\frac{1}{x}\right)^{6}+12\left(\frac{1}{x}\right)^{4}+60\left(\frac{1}{x}\right)^{2}+160$
$+240(x)^{2}+192(x)^{4}+64(x)^{6}$
$\therefore $ Sum of coefficients
$=1+12+60+160+240+192+64$
$=729$