Evaluate
\[i^{14762} + i^{14763} + i^{14764} + i^{14765}.\]
We can take out a factor of $i^{14762}$ to get
\[i^{14762} + i^{14763} + i^{14764} + i^{14765} = i^{14762} (1 + i + i^2 + i^3).\]Since $i^2 = -1$ and $i^3 = -i,$
\[1 + i + i^2 + i^3 = 1 + i - 1 - i = 0.\]Therefore, the expression is equal to $\boxed{0}.$
0
Level 1
Intermediate Algebra
Problem: Evaluate
\[i^{14762} + i^{14763} + i^{14764} + i^{14765}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To evaluate the expression, we first recognize that \(i\) is the imaginary unit defined as \(i = \sqrt{-1}\). The powers of \(i\) cycle every four terms: 

- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

This cycle repeats, so to find \(i^{14762}\), \(i^{14763}\), \(i^{14764}\), and \(i^{14765}\), we need to determine the remainder of each exponent when divided by 4.

Calculating the remainders:
- \(14762 \mod 4 = 2\) (since \(14762 = 4 \times 3690 + 2\))
- \(14763 \mod 4 = 3\) (since \(14763 = 4 \times 3690 + 3\))
- \(14764 \mod 4 = 0\) (since \(14764 = 4 \times 3691 + 0\))
- \(14765 \mod 4 = 1\) (since \(14765 = 4 \times 3691 + 1\))

Now substituting these remainders back into the cycle:
- \(i^{14762} = i^2 = -1\)
- \(i^{14763} = i^3 = -i\)
- \(i^{14764} = i^0 = 1\)
- \(i^{14765} = i^1 = i\)

Next, we sum these results:
\[
i^{14762} + i^{14763} + i^{14764} + i^{14765} = -1 + (-i) + 1 + i
\]

Grouping the real and imaginary parts:
- The real parts: \(-1 + 1 = 0\)
- The imaginary parts: \(-i + i = 0\)

Thus, the total sum is:
\[
0 + 0 = 0
\]

Therefore, the final answer is \(0\).<|END_OF_LATENT|> The final answer is $0$.