Question

please give me the answer of 5.i , iii, iv​

Answer 1

5.

i) 3n - 1

• taking n = 1 ; 3n-1 = 3-1 = 2
• taking n =2; 3n-2 = 6-1 = 5
• taking n =3; 3n-2 = 9-1 = 8
• taking n =4; 3n-2 = 12-1 = 11

So first 4 no of series are : 2,5,8,11

iii) 4n+3

• taking n =1; 4n+3 = 7
• taking n =2; 4n+3 = 11
• taking n =3; 4n+3 = 15
• taking n =4; 4n+3 = 19

So first 4 no of series are : 7,11,15,19

iv) n^2 + 1

• taking n =1; n^2 + 1 = 2
• taking n =2; n^2 + 1 = 5
• taking n =3; n^2 + 1 = 10
• taking n =4; n^2 + 1 = 17

So first 4 no of series are : 2 , 5 , 1 0 , 17