The first term of a series consisting of six terms is $m^0$. 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. (Here, $m$ is a natural number.)

If another series starts with $1.5$ and follows the same pattern of multiplication as the original series, what is the fourth term of this new series?

Question

The first term of a series consisting of six terms is $m^0$. 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. (Here, $m$ is a natural number.)

If another series starts with $1.5$ and follows the same pattern of multiplication as the original series, what is the fourth term of this new series?

Accepted Answer

**Solution:**

1. First, analyze the original series:
   - First term: $m^0 = 1$ (since any natural number to the power 0 is 1)
   - Second term: smallest even prime number = 2
   - Third term: cube of second term = $2^3 = 8$
   - Fourth term: square of third term = $8^2 = 64$
   - Fifth term: 16 times the fourth term = $16 	imes 64 = 1024$

So, the original series is: 1, 2, 8, 64, 1024, ...

2. Notice the pattern in terms of powers of 2:
   - $1 = 2^0$
   - $2 = 2^1$
   - $8 = 2^3$
   - $64 = 2^6$
   - $1024 = 2^{10}$

The exponents increase as 0, 1, 3, 6, 10, which are triangular numbers.

3. Now, for the new series starting with 1.5, the pattern of multiplication between terms is the key:
   - From 1 to 2, multiplication is $2$
   - From 2 to 8, multiplication is $4$
   - From 8 to 64, multiplication is $8$
   - From 64 to 1024, multiplication is $16$

4. Applying the same multiplication pattern to the new series starting at 1.5:
   - First term: $1.5$
   - Second term: $1.5 	imes 2 = 3$
   - Third term: $3 	imes 4 = 12$
   - Fourth term: $12 	imes 8 = 96$

**Answer:** $96$

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