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Q. The coefficient of $x^{6}$ in the expansion of $\left(1 + x + x^{2}\right)^{6}$ is

NTA AbhyasNTA Abhyas 2020Binomial Theorem

Solution:

Coefficient of $x^{6}$ in $\left(1 + x \left(1 + x\right)\right)^{6}$
$=1+{ }^{6} C_{1} x(1+x)+{ }^{6} C_{2} x^{2}(1+x)^{2}+{ }^{6} C_{3} x^{3}(1+x)^{3}+{ }^{6} C_{4} x^{4}(1+x)^{4}$
$\Rightarrow $ The coefficient of $x^{6}$ is
$\_{}^{6}C_{3}\times \_{}^{3}C_{3}+\_{}^{6}C_{4}\times \_{}^{4}C_{2}+\_{}^{6}C_{5}\times \_{}^{5}C_{1}+\_{}^{6}C_{6}\times \_{}^{6}C_{0}$
$=\frac{6 \times 5 \times 4}{3 \times 2}\times 1+\frac{6 \times 5}{2}\times \frac{4 \times 3}{2}+6\times 5+1\times 1$
$=20+90+30+1=141$