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The coefficient of $x^5$ in the expansion of $(1+x)^{21}$ + $(1+x)^{22}$ + ...... + $(1+x)^{30}$ is

$\begin{array}{1 1} =^{30}C_5 +^{20} C_5 \\ =^{51} C_5 \\ =^{31} C_6 -^{21} C_6 \\ =^{30}C_5 -^{20} C_5 \end{array} $

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1 Answer

$(1+x)^{21}+(1+x)^{22}+.........(1+x)^{30}$
$=(1+x)^{21}.\big(1+(1+x)+(1+x)^2+........((1+x)^9\big)$
$=(1+x)^{21}\bigg\{\large\frac{(1+x)^{10}-1}{(1+x)-1}\bigg\}$
$=\large\frac{1}{x}$$\big[(1+x)^{31}-(1+x)^{21}\big]$
$\therefore\:$ Coeff. of $x^5$ in this expansion=
Coeff. of $x^6$ in the expansion $\big[(1+x)^{31}-(1+x)^{21}\big]$
$=^{31}C_6-^{21}C_6$
answered Oct 14, 2013 by rvidyagovindarajan_1
 

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