The number of real negative terms in the binomial expansion of (1+i x)4 n-2, n ∈ N, x 0 is

Question

The number of real negative terms in the binomial expansion of (1+i x)4 n-2, n ∈ N, x >0 is (A) n (B) n+1 (C) n-1 (D) 2 n

Tardigrade Admin · Accepted Answer

Tr+1= 4 n-2 Cr(i x)r Tr+1 is negative, if ir is negative and real. ir=-1 ⇒ r=2,6,10, ldots which form an A.P. 0 ≤ r ≤ 4 n-2 4 n-2=2+(r-1) 4 ⇒ r=n The required number of terms is n.

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

$n$

$n + 1$

$n - 1$

$2 n$

$T_{r + 1} =^{4 n - 2} C_{r} (i x)^{r}$
$T_{r + 1}$ is negative, if $i^{r}$ is negative and real.
$i^{r} = - 1$
$\Rightarrow r = 2, 6, 10, \dots$ which form an A.P.
$0 \leq r \leq 4 n - 2$
$4 n - 2 = 2 + (r - 1) 4$
$\Rightarrow r = n$
The required number of terms is $n$ .

Q. The number of real negative terms in the binomial expansion of (1+ix)4n−2,n∈N,x>0 is

n

n+1

n−1

2n

Tr+1​=4n−2Cr​(ix)r Tr+1​ is negative, if ir is negative and real. ir=−1 ⇒r=2,6,10,… which form an A.P. 0≤r≤4n−2 4n−2=2+(r−1)4 ⇒r=n The required number of terms is n.

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

$n$

$n + 1$

$n - 1$

$2 n$

$T_{r + 1} =^{4 n - 2} C_{r} (i x)^{r}$
$T_{r + 1}$ is negative, if $i^{r}$ is negative and real.
$i^{r} = - 1$
$\Rightarrow r = 2, 6, 10, \dots$ which form an A.P.
$0 \leq r \leq 4 n - 2$
$4 n - 2 = 2 + (r - 1) 4$
$\Rightarrow r = n$
The required number of terms is $n$ .