Write the first five terms of the geometric sequence if the nth term is given by 36(1/3)^n-1

G1=36,G2=12,G3=4,G4=4/3,G5=4/9

Step-by-step explanation:

Since the nth term is given by;

Gn= 36(1/3)^n-1, then, substitute the values of n as 1,2,3,4 and 5 to get the values of G1,G2,G3,G4 and G5 respectively.

G1=36(1/3)^1-1 = 36

G2= 36(1/3)^2-1= 12

G3= 36(1/3)^3-1 = 4

G4= 36(1/3)^4-1 = 4/3

G5= 36(1/3)^5-1=4/9

36, 12, 4, 4/3, 4/9.

Step-by-step explanation:

We find each term by substituting the sequence number.

So the first term (where n = 1)

= 36(1/3)^(1-1)

=  36(1/3)^0

=  36 * 1

= 36.  

The second term is 36(1/3)^(2-1)  = 36 * 1/3

= 12.

We can  find subsequent terms by just multiplying the previous term by 1/3 so the third term = 12 * 1/3 = 4.

So the first 5 terms are   36, 12, 4, 4/3, 4/9.

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry