Question

The first negative term in the expansion \(\sqrt{(1+2x)^7}\) is the:

• A. 4th term
• B. 5th term
• C. 6th term
• D. 7th term

Hint:

Expand \((1+2x)^{7/2}\) as a binomial series and find the first value of r where the coefficient factor n-r+1 turns negative.

Answer 1

The Correct Option is C

Step 1: Set up the expansion.
We need the first negative term in the expansion of \(\sqrt{(1+2x)^7}\), which is the same as \((1+2x)^{7/2}\). Here the power \(n=\frac{7}{2}\) is not a whole number, so the binomial series does not stop after a fixed number of terms. Take \(x>0\), so \((2x)^r\) stays positive for every term, and the sign of each term depends only on its binomial coefficient.

Step 2: Write the general term.
For \((1+2x)^n\) with \(n=\frac{7}{2}\), the term at position \(r\) (counting the first term as \(r=0\)) is
\[ T_{r+1} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}(2x)^r \]
The numerator has \(r\) factors: \(n, n-1, n-2, \ldots, n-r+1\). The term first turns negative exactly when a new factor in this chain becomes negative for the first time.

Step 3: Track the sign of each factor.
With \(n=\frac{7}{2}=3.5\), the successive factors are
\[ 3.5,\ 2.5,\ 1.5,\ 0.5,\ -0.5,\ -1.5,\ldots \]
The first four of these (\(3.5, 2.5, 1.5, 0.5\)) are all positive, so the terms using \(r=0,1,2,3,4\) (that is, terms \(T_1\) to \(T_5\)) stay positive. The fifth factor is \(3.5-4=-0.5\), which is negative, and it enters the product when \(r=5\).

Step 4: Identify the term number.
Since \(r=5\) gives the first negative coefficient and the term number is \(r+1\), the first negative term is \(T_6\), the 6th term. Options (A) 4th term and (B) 5th term use \(r=3\) and \(r=4\), where every factor so far is still positive, so those terms are positive, not negative. Option (D) 7th term uses \(r=6\), which comes after the sign change at \(r=5\), so it is not the first negative term.

Final Answer:
The first negative term is the 6th term.
\[ \boxed{\text{6th term}} \]