Question

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

• A. i
• B. 1
• C. 0
• D. -i

Answer 1

Answer: (C)

0

Answer 2

The given series is a geometric series: $1+i+i^2+\dots$ to 1000 terms.

Here, the first term is $a = 1$.
The common ratio is $r = i$.
The number of terms is $n = 1000$.

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

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

Now, we need to evaluate $i^{1000}$. We know that the powers of $i$ follow a cycle of 4:
$i^1 = i$
$i^2 = -1$
$i^3 = i^2 \cdot i = -i$
$i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$

To find $i^{1000}$, we divide the exponent by 4:
$1000 \div 4 = 250$ with a remainder of 0.
This means $i^{1000} = (i^4)^{250} = 1^{250} = 1$.

Substitute $i^{1000} = 1$ back into the sum equation:
$S_{1000} = \frac{1(1-1)}{1-i}$
$S_{1000} = \frac{0}{1-i}$
$S_{1000} = 0$

Alternative Method (Grouping):

We know that the sum of any four consecutive powers of $i$ is zero:
$i^k + i^{k+1} + i^{k+2} + i^{k+3} = i^k(1 + i + i^2 + i^3)$
$= i^k(1 + i - 1 - i)$
$= i^k(0) = 0$

The given series is $1+i+i^2+\dots+i^{999}$.
There are 1000 terms in the series. Since 1000 is a multiple of 4 ($1000 = 4 \times 250$), we can group the terms into 250 sets of four consecutive powers of $i$:
$(1+i+i^2+i^3) + (i^4+i^5+i^6+i^7) + \dots + (i^{996}+i^{997}+i^{998}+i^{999})$

Each group sums to 0:
$(1+i-1-i) = 0$
$(i^4(1+i+i^2+i^3)) = 1(0) = 0$
And so on.

Since there are 250 such groups, and each group sums to 0, the total sum is:
$S_{1000} = 0 + 0 + \dots + 0$ (250 times)
$S_{1000} = 0$

Therefore, the sum is 0.