The coefficient of x^{10} in the expansion of | vedclass

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(A) The general term in the expansion of $(x + \frac{2}{x} - 5)^{12}$ is given by the multinomial theorem as: $\frac{12!}{a!b!c!} (x)^a (\frac{2}{x})^

Vedclass · Accepted Answer

$1674$

Vedclass · Answer

(A) The general term in the expansion of $(x + \frac{2}{x} - 5)^{12}$ is given by the multinomial theorem as: $\frac{12!}{a!b!c!} (x)^a (\frac{2}{x})^b (-5)^c$,where $a+b+c = 12$.This simplifies to $\frac{12!}{a!b!c!} 2^b (-5)^c x^{a-b}$.We want the coefficient of $x^{10}$,so $a-b = 10$.Substituting $a = b+10$ into $a+b+c = 12$,we get $(b+10) + b + c = 12$,which implies $2b + c = 2$.Possible non-negative integer solutions for $(a, b, c)$ are:$1$) If $b=0$,then $c=2$ and $a=10$. The term is $\frac{12!}{10!0!2!} (2)^0 (-5)^2 = 66 	imes 25 = 1650$.$2$) If $b=1$,then $c=0$ and $a=11$. The term is $\frac{12!}{11!1!0!} (2)^1 (-5)^0 = 12 	imes 2 = 24$.Summing these,the coefficient is $1650 + 24 = 1674$.

The coefficient of $x^{10}$ in the expansion of $(x+\frac{2}{x}-5)^{12}$ is

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