Question

The cube of first 5 natural numbers of the form 3n + 2, where 'n' is a natural number, will be of the form _______.

• A. 3n
• B. 3n + 2
• C. 3n - 2
• D. 3n +1

Answer 1

The correct option is B 3n + 2

The first 5 natural numbers of the form 3n + 2 are as follows:
n = 1; 3(1) + 2= 5
n =2; 3(2) + 2 = 8
n = 3; 3(3) + 2= 11
n =4; 3(4) + 2 =14
n = 5; 3(5) + 2= 17

The cubes of the natural numbers which are of the form 3n + 2 are:
$5^3=5\times 5\times 5=1258^3=8\times 8\times 8=512{11}^3=11\times 11\times 11=1331{14}^3=14\times 14\times 14=2744{17}^3=17\times 17\times 17=4913$

Verification:
$125=3\times 41+2512=3\times 170+21331=3\times 443+22744=3\times 914+24913=3\times 1637+2$

The cube of first 5 natural numbers of the form 3n + 2 will be of the form 3n + 2.