Question

The number of real negative terms in the binomial expansion of ${(1+ix)}^{4n+2}$, $n\in N,x>0$, where $i=\sqrt{-1}$, is-

• A. $n-1$
• B. $n$
• C. $n+1$
• D. $2n$

Answer 1

Answer: (C)

$n+1$

$T_{r+1}={{}^{4n+2}C}_r{i}^{.r}{x}^r$

$\because$ $x>0$, $T_{r+1}$ is negative real when $i^r$ is negative and real

$\Rightarrow$ $i^{.r}=-1$
$\Rightarrow$ $r=2,6,10,....$ which forms an A.P with $k^{th}$term $4k-2$

Also $0\le r\le 4n+2$

$\therefore$ $4k-2=4n+2$
$\Rightarrow$ $k=n+1$