What are the first four terms of 3n²

3, 12, 27, 48

Step-by-step explanation:

To find the first 4 terms substitute n = 1, 2, 3, 4 into the rule, that is

n = 1 → 3 × 1² = 3 × 1 = 3 ← first term

n = 2 → 3 × 2² = 3 × 4 = 12 ← second term

n = 3 → 3 × 3² = 3 × 9 = 27 ← third term

n = 4 → 3 × 4² = 3 × 16 = 48 ← fourth term

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: