A sequence of number u1,u2,u3.. satisfies the relation (3n-2) Un +1 =(3n+1) Un, for all positive intergers n, if U1=1.find Un+1=U3 and u4
To find U3, we substitute n=1 into the given relation:
(3(1)-2) U2 + 1 = (3(1)+1) U1
U2 = 2U1 = 2(1) = 2
Then, substituting n=2 into the given relation, we can find U3:
(3(2)-2) U3 + 1 = (3(2)+1) U2
4U3 + 1 = 7(2)
4U3 = 13
U3 = 13/4
To find U4, we substitute n=3 into the given relation:
(3(3)-2) U4 + 1 = (3(3)+1) U3
7U4 + 1 = 10(13/4)
7U4 = 123/4
U4 = 123/28
Therefore, U3 = 13/4 and U4 = 123/28.
To find the value of Un+1 when Un = U3, we can insert the values into the given relation and solve for Un+1.
Given:
Un = U3
U1 = 1
Using the relation (3n-2)Un + 1 = (3n+1)Un, let us substitute n with 3:
(3n-2)Un + 1 = (3n+1)Un
(3*3-2)U3 + 1 = (3*3+1)U3
(7)U3 + 1 = (10)U3
7U3 + 1 = 10U3
Now, let's solve for U3:
7U3 + 1 = 10U3
Subtract 7U3 from both sides:
1 = 3U3
Divide both sides by 3:
1/3 = U3
So, U3 = 1/3.
Now, let's find U4 by substituting n with 4:
(3n-2)Un + 1 = (3n+1)Un
(3*4-2)U4 + 1 = (3*4+1)U4
(10)U4 + 1 = (13)U4
To solve for U4, we can subtract (10)U4 from both sides:
1 = 3U4 - 10U4
Combine like terms:
1 = -7U4
Divide both sides by -7:
-1/7 = U4
So, U4 = -1/7.