Let K be the coefficient of x^4 in the expans | vedclass

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(B) The expansion is $(1 + x + ax^2)^{10} = \sum_{i+j+k=10} \frac{10!}{i!j!k!} (1)^i (x)^j (ax^2)^k = \sum \frac{10!}{i!j!k!} a^k x^{j+2k}$.We want th

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$-4$

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(B) The expansion is $(1 + x + ax^2)^{10} = \sum_{i+j+k=10} \frac{10!}{i!j!k!} (1)^i (x)^j (ax^2)^k = \sum \frac{10!}{i!j!k!} a^k x^{j+2k}$.We want the coefficient of $x^4$,so $j+2k = 4$ with $i+j+k = 10$.Possible non-negative integer solutions $(i, j, k)$:$1$. If $k=0$,then $j=4$,$i=6$. Coefficient $= \frac{10!}{6!4!0!} a^0 = 210$.$2$. If $k=1$,then $j=2$,$i=7$. Coefficient $= \frac{10!}{7!2!1!} a^1 = 360a$.$3$. If $k=2$,then $j=0$,$i=8$. Coefficient $= \frac{10!}{8!0!2!} a^2 = 45a^2$.Thus,$K(a) = 45a^2 + 360a + 210$.To minimize $K(a)$,we find the vertex of the parabola $f(a) = 45a^2 + 360a + 210$.The minimum occurs at $a = -\frac{b}{2A} = -\frac{360}{2(45)} = -\frac{360}{90} = -4$.

Let $K$ be the coefficient of $x^4$ in the expansion of $(1 + x + ax^2)^{10}$. What is the value of $a$ that minimizes $K$?

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