Question

If the third and fourth terms in the expansion \( (2x + \frac{1}{8})^{10} \) are equal, then the value of \( x \) is

• A. \( \frac{1}{8} \)
• B. \( \frac{2}{7} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{8}{3} \)

Hint:

When consecutive terms are equal in binomial expansion, divide one term by the previous to simplify quickly.

Answer 1

The Correct Option is C

Step 1: Write the general term.
The general term in expansion of \( (a+b)^n \) is:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r. \]
Here:
\[ a=2x,\quad b=\frac{1}{8},\quad n=10. \]

Step 2: Write third and fourth terms.
Third term corresponds to \(r=2\):
\[ T_3 = \binom{10}{2}(2x)^8\left(\frac{1}{8}\right)^2. \]
Fourth term corresponds to \(r=3\):
\[ T_4 = \binom{10}{3}(2x)^7\left(\frac{1}{8}\right)^3. \]

Step 3: Set the terms equal.
\[ \binom{10}{2}(2x)^8\left(\frac{1}{8}\right)^2 = \binom{10}{3}(2x)^7\left(\frac{1}{8}\right)^3. \]

Step 4: Simplify coefficients.
\[ \binom{10}{2}=45,\quad \binom{10}{3}=120. \]
So,
\[ 45(2x)^8\left(\frac{1}{8}\right)^2 = 120(2x)^7\left(\frac{1}{8}\right)^3. \]

Step 5: Cancel common terms.
Divide both sides by \( (2x)^7\left(\frac{1}{8}\right)^2 \):
\[ 45(2x)=120\left(\frac{1}{8}\right). \]

Step 6: Solve for \(x\).
\[ 90x=\frac{120}{8}=15. \]
\[ x=\frac{15}{90}=\frac{1}{6}. \]

Step 7: Final conclusion.
\[ \boxed{\frac{1}{6}} \]