First slide

Binomial theorem for positive integral Index

Question

The coefficient of $x^{4}$ in the expansion ${(1 + x + x^{2} + x^{3})}^{6}$ in power of $x$ is

Moderate

A

120
B

130
C

110
D

150

Solution

We have , ${(1 + x + x^{2} + x^{3})}^{6} = (1 + x)^{6} {(1 + x^{2})}^{6}$
$\begin{array}{l} = (\sum_{r = 0}^{6}^{6} C_{r} x^{5}) (\sum_{s = 0}^{6}^{6} C_{s} x^{2 s}) \\ = (^{6} C_{0} +^{6} C_{1} x +^{6} C_{2} x^{2} +^{6} C_{3} x^{3} +^{6} C_{4} x^{4} + \dots) x (^{6} C_{0} +^{6} C_{1} x^{2} +^{6} C_{2} x^{4} + \dots) \end{array}$
$∴$ Coefficient of $x^{4}$ in ${(1 + x + x^{2} + x^{3})}^{6} = (^{6} C_{0} \times^{6} C_{2} +^{6} C_{2} \times^{6} C_{1} +^{6} C_{4} \times^{6} C_{0}) = 15 + 90 + 15 = 120$

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