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: Rewrite the root as a fractional power.
$\sqrt{(1+2x)^7} = (1+2x)^{7/2}$. Take $n=7/2$ and expand using the general binomial series $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3+\cdots$, with $y=2x$.

Step 2: Compute each binomial coefficient as a fraction.
Using $C(n,r) = C(n,r-1)\times\frac{n-r+1}{r}$, starting from $C(7/2,0)=1$:
$C(7/2,1) = 1 \times \frac{7/2}{1} = \frac{7}{2}$
$C(7/2,2) = \frac{7}{2}\times\frac{5/2}{2} = \frac{35}{8}$
$C(7/2,3) = \frac{35}{8}\times\frac{3/2}{3} = \frac{35}{16}$
$C(7/2,4) = \frac{35}{16}\times\frac{1/2}{4} = \frac{35}{128}$
$C(7/2,5) = \frac{35}{128}\times\frac{-1/2}{5} = -\frac{7}{256}$

Step 3: Spot where the sign flips.
All the coefficients up to $C(7/2,4)$ come out positive: $1, \frac{7}{2}, \frac{35}{8}, \frac{35}{16}, \frac{35}{128}$. The very next one, $C(7/2,5) = -\frac{7}{256}$, is negative. Since $(2x)^r$ does not change sign when $x>0$, this is exactly the point where the term itself turns negative.

Step 4: Match the coefficient index to the term number.
$C(7/2,r)$ belongs to term number $r+1$, so $r=5$ gives the 6th term. This confirms the first negative term is the 6th term, so options (A), (B) and (D) do not fit.

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