The coefficient of x^5 in (3+x+x^2)^6 is | vedclass

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(D) The general term in the expansion of $(3+x+x^2)^6$ is given by the multinomial theorem as $\frac{6!}{p!q!r!} 3^p \cdot x^q \cdot (x^2)^r = \frac{6

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$2178$

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(D) The general term in the expansion of $(3+x+x^2)^6$ is given by the multinomial theorem as $\frac{6!}{p!q!r!} 3^p \cdot x^q \cdot (x^2)^r = \frac{6!}{p!q!r!} 3^p \cdot x^{q+2r}$,where $p+q+r=6$.We need the coefficient of $x^5$,so we set $q+2r=5$,which implies $q=5-2r$.Substituting $q$ into $p+q+r=6$,we get $p+(5-2r)+r=6$,which simplifies to $p=1+r$.Since $p, q, r \ge 0$,we test possible values for $r$:$1$) If $r=0$,then $p=1$ and $q=5$. The term is $\frac{6!}{1!5!0!} 3^1 = 6 	imes 3 = 18$.$2$) If $r=1$,then $p=2$ and $q=3$. The term is $\frac{6!}{2!3!1!} 3^2 = \frac{720}{2 	imes 6} 	imes 9 = 60 	imes 9 = 540$.$3$) If $r=2$,then $p=3$ and $q=1$. The term is $\frac{6!}{3!1!2!} 3^3 = \frac{720}{6 	imes 2} 	imes 27 = 60 	imes 27 = 1620$.Summing these coefficients,we get $18 + 540 + 1620 = 2178$.

The coefficient of $x^5$ in $(3+x+x^2)^6$ is

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