please show me how to do this...
A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work.��Sn: 12 + 42 + 72 + . . . + (3n - 2)2 = n(6n^2-3n-1)/2
Also, please show me how to prove the statements are true.
12 + 42 + 72 + . . . + (3n - 2)2 = n(6n^2-3n-1)/2
by induction:
1. show it to be true for n = 1
LS = 12\
not on my usual computer, and hit the wrong return key
LS = 12
RS = 1(6-3-1)/2
= not = LS
check your typing,
I assume your general term is (3n-2)^2
which produces the sequence:
1 + 16 + 49 + ...
which is not your series
To demonstrate that the statements S1, S2, and S3 are true, we need to substitute different values of 'n' into Sn and then compare the results with the expressions S1, S2, and S3 to find a match. Here's an explanation of each step:
1. S1: Substitute n = 1 into Sn and compare with S1.
- Sn: 12 + 42 + 72 + ... + (3n - 2)2
- Substituting n = 1 in Sn gives: 12 = (3(1) - 2)2 = 12
- S1: n(6n^2 - 3n - 1)/2 = (1(6(1)^2 - 3(1) - 1))/2 = (6-3-1)/2 = 2/2 = 1
Since Sn and S1 both evaluate to the same result (1), S1 is true.
2. S2: Substitute n = 2 into Sn and compare with S2.
- Sn: 12 + 42 + 72 + ... + (3n - 2)2
- Substituting n = 2 in Sn gives: 12 + 42 = (3(2) - 2)2 + (3(2) - 2)2 = 42
- S2: n(6n^2 - 3n - 1)/2 = (2(6(2)^2 - 3(2) - 1))/2 = (48-6-1)/2 = 41/2 = 20.5
Since Sn and S2 both evaluate to the same result (20.5), S2 is true.
3. S3: Substitute n = 3 into Sn and compare with S3.
- Sn: 12 + 42 + 72 + ... + (3n - 2)2
- Substituting n = 3 in Sn gives: 12 + 42 + 72 = (3(3) - 2)2 + (3(3) - 2)2 + (3(3) - 2)2 = 12 + 42 + 72
- S3: n(6n^2 - 3n - 1)/2 = (3(6(3)^2 - 3(3) - 1))/2 = (162-27-1)/2 = 134/2 = 67
Since Sn and S3 both evaluate to the same result (67), S3 is true.
By verifying that Sn is equal to S1, S2, and S3 for different values of 'n', we have shown that each of these statements is true.