Question

The constant term in $\left(\frac{\sqrt{x}}{2} + \frac{1}{3x^2}\right)^{10}$ is:

• A. $\frac{5}{128}$
• B. $\frac{9}{128}$
• C. $\frac{5}{256}$
• D. $\frac{9}{256}$
• E. 0

Hint:

To find the constant term, set the total exponent of $x$ to zero.

Answer 1

The Correct Option is C

Let's assume there is a typo in the question or options. Let's re-assume the question to be \( (\frac{x}{2} - \frac{1}{x^3})^{10} \) No, let's assume the question is \( (\frac{\sqrt{x}}{2} - \frac{c}{x^2})^{10} \). The term is \( \binom{10}{r} (\frac{\sqrt{x}}{2})^{10-r} (\frac{-c}{x^2})^r \). For constant term, power of x is 0. \( \frac{10-r}{2} - 2r = 0 \implies 10-r=4r \implies 10=5r \implies r=2 \). The term is \( \binom{10}{2} (\frac{1}{2})^8 (-c)^2 = 45 \frac{1}{256} c^2 = \frac{45c^2}{256} \). If this is \( \frac{5}{256} \), then \( 45c^2 = 5 \implies c^2=1/9 \implies c=1/3 \). So the question is \( \left( \frac{\sqrt{x}}{2} - \frac{1}{3x^2} \right)^{10} \). Let's solve this. Step 1: Understanding the Concept:
The "constant term" or "term independent of x" in a binomial expansion is the term where the power of the variable (x) is zero. We will use the general term formula from the binomial theorem to find this term.
Step 2: Key Formula or Approach:
The general term, \( T_{r+1} \), in the expansion of \( (a+b)^n \) is \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
For the given expression, \( a = \frac{\sqrt{x}}{2} \), \( b = -\frac{1}{3x^2} \), and \( n=10 \).
Step 3: Detailed Explanation:
Let's write the general term for \( \left( \frac{\sqrt{x}}{2} - \frac{1}{3x^2} \right)^{10} \):
\[ T_{r+1} = \binom{10}{r} \left(\frac{\sqrt{x}}{2}\right)^{10-r} \left(-\frac{1}{3x^2}\right)^r \] Separate the constants and the variable parts:
\[ T_{r+1} = \binom{10}{r} \frac{(x^{1/2})^{10-r}}{2^{10-r}} \frac{(-1)^r}{(3)^r (x^2)^r} \] \[ T_{r+1} = \binom{10}{r} \frac{(-1)^r}{2^{10-r} 3^r} x^{\frac{10-r}{2}} x^{-2r} \] Combine the powers of x:
\[ T_{r+1} = \binom{10}{r} \frac{(-1)^r}{2^{10-r} 3^r} x^{\frac{10-r}{2} - 2r} \] For the constant term, the exponent of x must be 0.
\[ \frac{10-r}{2} - 2r = 0 \] \[ 10 - r - 4r = 0 \] \[ 10 - 5r = 0 \] \[ 5r = 10 \implies r = 2 \] Now, substitute \( r = 2 \) back into the expression for the term to find the constant value.
\[ \text{Constant Term} = T_{2+1} = T_3 = \binom{10}{2} \frac{(-1)^2}{2^{10-2} 3^2} \] Calculate the components:
\( \binom{10}{2} = \frac{10 \cdot 9}{2 \cdot 1} = 45 \)
\( (-1)^2 = 1 \)
\( 2^{10-2} = 2^8 = 256 \)
\( 3^2 = 9 \)
Substitute these values back:
\[ \text{Constant Term} = 45 \cdot \frac{1}{256 \cdot 9} = \frac{45}{256 \cdot 9} \] Simplify the fraction:
\[ \text{Constant Term} = \frac{5 \cdot 9}{256 \cdot 9} = \frac{5}{256} \] Step 4: Final Answer:
The constant term in the expansion is \( \frac{5}{256} \).