2. Section 7.2
Sign In
n | t(n) |
---|---|
1 | 12 |
2 | 36 |
3 | 108 |
4 | 324 |
Dwayne:
n | t(n) |
---|---|
1 | 12 |
2 | 36 |
3 | 108 |
4 | 324 |
Observation: The tables are the same.
n | t(n) |
---|---|
1 | 10.3 |
2 | 11.5 |
3 | 12.7 |
4 | 13.9 |
Dwayne:
n | t(n) |
---|---|
1 | 10.3 |
2 | 11.5 |
3 | 12.7 |
4 | 13.9 |
Are Both Forms Correct? Yes.
n | 4(3)^n | t(n) |
---|---|---|
1 | 4(3)^1 | 12 |
2 | 4(3)^2 | 36 |
3 | 4(3)^3 | 108 |
4 | 4(3)^4 | 324 |
Let's also calculate the first four terms for Dwayne.
n | 12(3)^(n-1) | t(n) |
---|---|---|
1 | 12(3)^(1-1) | 12 |
2 | 12(3)^(2-1) | 36 |
3 | 12(3)^(3-1) | 108 |
4 | 12(3)^(4-1) | 324 |
As we can see, both equations give the same sequence. Therefore, both forms of the equation are correct.
First term:& t(n)= a_1b^(n-1) & a_1= first term [0.8em] Zeroth term:& t(n)= a_0b^n & a_0= zeroth term As we can see, Dwayne has written the equation in first-term form.
n | 9.1+1.2n | t(n) |
---|---|---|
1 | 9.1+1.2( 1) | 10.3 |
2 | 9.1+1.2( 2) | 11.5 |
3 | 9.1+1.2( 3) | 12.7 |
4 | 9.1+1.2( 4) | 13.9 |
Let's also calculate the first four terms for Dwayne.
n | 10.3+1.2(n-1) | t(n) |
---|---|---|
1 | 10.3+1.2( 1-1) | 10.3 |
2 | 10.3+1.2( 2-1) | 11.5 |
3 | 10.3+1.2( 3-1) | 12.7 |
4 | 10.3+1.2( 4-1) | 13.9 |
As we can see, both equations give the same sequence. Therefore, both forms of the equation are correct.
first termform, because his equation starts with the first term of the sequence. We have to subtract 1 from n because if we substitute n=1 into the
first termform we are left with a_1.