If i2=-1 and displaystyle ∑ r = 1n(i)r,∀ n∈ N, is a non-zero real number, then n can be

Question

If i2=-1 and displaystyle ∑ r = 1n(i)r,∀ n∈ N, is a non-zero real number, then n can be (A) 100 (B) 201 (C) 302 (D) 403

Tardigrade Admin · Accepted Answer

ip+ip+1+ip+2+ip+3=0, ∀ p ∈ N So ∑r=1n(i)r= beginarraycc0 text for n=4 k i text for n=4 k+1 i-1 text for n=4 k+2 -1 text for n=4 k+3 endarray ∀ k ∈ N. Hence, n can be 403 for the sum to be non-zero real.

Q. If $i^{2} = - 1$ and $r = 1 \sum n (i)^{r}, \forall n \in N,$ is a non-zero real number, then $n$ can be

$100$

$201$

$302$

$403$

$i^{p} + i^{p + 1} + i^{p + 2} + i^{p + 3} = 0, \forall p \in N$
So $\sum_{r = 1}^{n} (i)^{r} = ⎩ ⎨ ⎧ 0 i i - 1 - 1 for n = 4 k for n = 4 k + 1 for n = 4 k + 2 for n = 4 k + 3 \forall k \in N$
Hence, $n$ can be $403$ for the sum to be non-zero real.

Q. If i2=−1 and r=1∑n​(i)r,∀n∈N, is a non-zero real number, then n can be

100

201

302

403