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The sum of the coefficients of integral powers of $x$ in the expansion of $(1+\sqrt x)^{40}$ is

$\begin{array}{1 1} 0 \\ 1 \\ \frac{1}{2}\\ \frac{3}{2} \end{array} $

1 Answer

$(1+\sqrt x)^{40}=^{40}C_0+^{40}C_1\sqrt x+^{40}C_2x+^{40}C_3(\sqrt x)^3+......^{40}C_{40}x^{20}$
$=(^{40}C_0+^{40}C_2x+...^{40}C_{40}x^{20})+(^{40}C_1\sqrt x+^{40}C_3(\sqrt x)^3+....^{40}C_{39}(\sqrt x)^{39})$
$=$(sum of integral power of $x$ terms)+(sum of non integral power terms)
By taking $x=1$ and`$ \sqrt x=1$ and then again $\sqrt x =-1$
in the above expansion we get
$1=(^{40}C_0+^{40}C_2+...^{40}C_{40})+(^{40}C_1+^{40}C_3+....^{40}C_{39})$....(i)
and
$0=(^{40}C_0+^{40}C_2+...^{40}C_{40})-(^{40}C_1+^{40}C_3+....^{40}C_{39})$...(ii)
Adding (i) and (ii)
$1=2(^{40}C_0+^{40}C_2+..........^{40}C_{40})$
$\Rightarrow\:^{40}C_0+^{40}C_2+..........^{40}C_{40}=\large\frac{1}{2}$
answered Oct 22, 2013 by rvidyagovindarajan_1
 

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