Question

If $i^2 = -1$, then sum $1 + i^2 + i^4 + ...$ to 1000 terms is equal to

• A. 0 1
• B. 10 1
• C. 0 -1
• D. 0 0

Answer 1

Answer: (D)

0

Answer 2

The given series is $1 + i^2 + i^4 + ...$ to 1000 terms.
Let's analyze the terms of the series:
The first term is $a_1 = 1$.
The second term is $a_2 = i^2 = -1$.
The third term is $a_3 = i^4 = (i^2)^2 = (-1)^2 = 1$.
The fourth term is $a_4 = i^6 = (i^2)^3 = (-1)^3 = -1$.

The series is $1, -1, 1, -1, ...$. This is an alternating series.

This is a geometric progression with:
First term, $a = 1$.
Common ratio, $r = \frac{i^2}{1} = i^2 = -1$.
Number of terms, $n = 1000$.

The sum of a finite geometric series is given by the formula:
$S_n = \frac{a(1-r^n)}{1-r}$

Substitute the values:
$S_{1000} = \frac{1(1-(-1)^{1000})}{1-(-1)}$

Now, evaluate $(-1)^{1000}$. Since 1000 is an even number, $(-1)^{1000} = 1$.

Substitute this back into the sum formula:
$S_{1000} = \frac{1(1-1)}{1+1}$
$S_{1000} = \frac{0}{2}$
$S_{1000} = 0$

Alternatively, we can group the terms:
The series is $1 + (-1) + 1 + (-1) + ...$ to 1000 terms.
We can group the terms in pairs:
$(1 + (-1)) + (1 + (-1)) + ...$
Each pair sums to $1 - 1 = 0$.
Since there are 1000 terms, there are $1000 / 2 = 500$ such pairs.
The total sum is $500 \times 0 = 0$.

The sum of the series is 0.