The first term of a series consisting of six terms is $m^0$, where $m$ is a natural number. The second term is the smallest even prime number. The third term is the cube of the second term. The fourth term is the square of the third term. The fifth term is 16 times the fourth term.

**Question:** Find the sixth term of the series.

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The first term of a series consisting of six terms is $m^0$, where $m$ is a natural number. The second term is the smallest even prime number. The third term is the cube of the second term. The fourth term is the square of the third term. The fifth term is 16 times the fourth term.

**Question:** Find the sixth term of the series.

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**Solution:**

1. The first term is $m^0$. Since $m$ is a natural number, any natural number raised to the power 0 equals 1. So, the first term is:
   $$m^0 = 1$$

2. The second term is the smallest even prime number, which is 2.

3. The third term is the cube of the second term:
   $$2^3 = 8$$

4. The fourth term is the square of the third term:
   $$8^2 = 64$$

5. The fifth term is 16 times the fourth term:
   $$16 	imes 64 = 1024$$

6. Observing the pattern, the terms can be rewritten as powers of 2:
   - First term: $1 = 2^0$
   - Second term: $2 = 2^1$
   - Third term: $8 = 2^3$
   - Fourth term: $64 = 2^6$
   - Fifth term: $1024 = 2^{10}$

The exponents follow the sequence: 0, 1, 3, 6, 10, ... which are triangular numbers.

7. The sixth triangular number is 15, so the sixth term is:
   $$2^{15} = 32768$$

**Answer:** $32768$

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