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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(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 $K$?

Similar Questions

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

If the coefficients of $x^9$ and $x^{10}$ in the binomial expansion of $(3+\frac{x}{2})^n$ are equal,then $n=$

The middle term in the expansion of ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$ in powers of $x$ is

The middle term in the expansion of ${\left( x + \frac{1}{x} \right)^{10}}$ is

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module