FIND THE Nth TERM USING "POWER" AND "FRACTIONS"

In this hub I intend to establish the relationship between "powers" and fractions in the n^th term of a sequence.

Complete the tables below given that the n^th term is:

(i) We will complete the following table where n^th term is 2ⁿ (2 to the power of ^"n")

This can be written as u_n = 2ⁿ

So lets start to substitute

n     1     2     3     4     5

u_n    2     4     8    16    32

         ↔     ↔    ↔    ↔ (Do you see a pattern)

          2      4      8     16 (The number is being doubled each time, this suggests that the rule involves 2 to the ^power of something.)

u_n = 2ⁿ

When ^"n" is 1... 2¹(2 to the power of ¹) = 2

when ^"n" is 2... 2²(2 to the power of ²) = 4

when ^"n" is 3... 2³(2 to the power of 3) = 8

when ^"n" is 4... 2⁴(2 to the power of 4) = 16

when ^"n" is 5... 2⁵(2 to the power of 5) = 32

 

 

 

(ii) Next we will complete the table where the n^th term is 2^{n - 1} + n (2 to the power of ^{"n -1"} + n)

This can be written as u_n = 2^{n - 1} + n

So lets start to substitute

n     1     2     3     4      5

u_n    2     4     7     12    21

u_n = 2^{n - 1} + n

When ^"n" is 1... 2^{1 - 1} = 2⁰ which = 1 + 1(n) = 2

When ^"n" is 2... 2^{2 - 1} =2¹ which = 2 + 2(n) = 4

When ^"n" is 3... 2^{3 - 1} = 2² which = 4 + 3(n) = 7

When ^"n" is 4... 2^{4 - 1} = 2³ which = 8 + 4(n) =12

When ^"n" is 5... 2^{5 - 1} = 2⁴ which =16 + 5(n) =21

♦ Note that ^"n^0" will always be 1

 

 

 

(iii) Now to complete the table where the n^th term is 4^{n - 2}(4 to the power of ^{"n - 2"})

This can be written as u_n = 4^{n - 2}

So lets start to substitute

n     1      2     3     4     5

u_n 0.25   1     4    16   64

u_n = 4^{n - 2}

When ^"n" is 1... 4^{1 - 2} = 4^-1 (remember to cross multiply here to get rid of the ^-1}

¹/4 x ^-1/1 ( these two 1's on right hand side cancel each other out, leaving ¼

which = 1 over 4(¼) = 0.25

When ^"n" is 2... 4^{2 - 2} = 4⁰ which = 1

When ^"n" is 3... 4^{3 - 2} = 4¹ which = 4

When ^"n" is 4... 4^{4 - 2} = 4² which = 16

When ^"n" is 5... 4^{5 - 2} = 4³ which = 64

 

 

 

Now to find the formula for u_n in the following tables

( i )

n     1     2     3     4     5

u_n    3   12    27   48   75

           ↔   ↔     ↔   ↔

            9   15     21   27 (no pattern yet, lets find the difference again)

              ↔     ↔    ↔

               6      6      6 (each one has a difference of 6),

can we see a pattern in the u_n , above since the second line has a constant number this suggests we are looking for n²

u_n = ?

u_n = n²

If we square the 1² that = 1, but we need 3 so we'll multiply by 3 ( 1 x 3 = 3)

Now to square the 2² that = 4, but we need 12 so again we multiply by 3 (4 x 3 = 12)

So we need to put 3 into the formula

u_n = 3 n²

we'll do the next one to see if it fits the table above

u_n = 3 n²

u_n = 3 x 3² = 3 x 9 = 27

u_n = 3 x 4² = 3 x 16 = 48

u_n = 3 x 5² = 3 x 25 = 75

so the formula is correct

u_n = 3 n²

 

 

( ii ) Example

n     1     2     3     4      5

u_n     2     8    26   80   242

          ↔     ↔    ↔     ↔

           6      18   54   162 (do you see a pattern in the difference?)

               ↔      ↔    ↔     (If not see second line)

               12      36   108  (they are multiples of 3, this suggests the rule

involves 3 ^{to the power} of something)

u_n = ?

u_n = 3ⁿ - 1

So following our rule which suggests 3^{to the power ,}

3¹ = 3 ( but if you look at the table we only need 2, so we'll take 1 away)

Our formula will now be 3ⁿ - 1

Now to do the rest to see that they comply with our formula.

3²= 9 - 1 = 8

3³ = 27 - 1 = 26

3⁴ = 81 - 1 = 80

3⁵ = 243 - 1 = 242

so the formula is correct

u_n = 3ⁿ - 1

 

 

 

( iii ) Next Example

n     1     2     3     4     5

u_n     3     6    11   20   37

           ↔    ↔   ↔    ↔

            3      5    9    17 ( no pattern yet,)

               ↔     ↔   ↔

                2      4     8 (each term is being doubled, so this suggests that the rule

involves 2^{to the power of} something)

u_n = ?

u_n = 2ⁿ

So following our rule which suggests 2^{to the power ,}

2¹ = 2 (but if you look at the table above you'll see we need 3, so we'll add 1 on)

Is our formula u_n = 2ⁿ + 1

Lets SEE!

2¹ = 2 + 1 = 3 (ok so far but look carefully)

Next one......

2² = 4 + 1 = 5 (this isn't going to work as we need 6, see above table)

So u_n = 2ⁿ + 1 is wrong!

so instead of adding on 1 we need to add "n"

Our formula now is u_n = 2ⁿ + n

Therefore;

2¹ = 2 + 1 = 3

2² = 4 + 2 = 6

2³ = 8 + 3 = 11

2⁴ = 16 + 4 = 20

2⁵ = 32 + 5 = 37

if you check back to the above table the formula is correct

u_n = 2ⁿ + n

 

 

 

Find the n^th term for each of the following sequences:

 

(i) 1, 2, 4, 8

Firstly we will put these values into a table

n     1     2     3     4

u_n     1     2     4     8

           ↔    ↔    ↔ (now to find the differences)

            1     2      4 (The number is being doubled each time, this suggests that the rule

involves 2 to the ^power of something.)

u_n = ?

u_n = 2ⁿ

This time the formula involves subtraction ^{of the power}

Therefore; u_n = 2^{n - 1}

2^{1 - 1} = n⁰ = 2⁰ which equals 1

2^{2 - 1} = n¹ = 2¹ which equals 2

2^{3 - 1} = n² = 2² which equals 4

2^{4 - 1} = n³ = 2³ which equals 8

if you check back to the above table the formula is correct

u_n = 2^{n - 1}

 

 

 

(ii) 8, 14, 32, 86

 

Lets put these values into a table

n     1     2     3     4

u_n     8   14    32   86

          ↔    ↔     ↔ ( now to find the differences)

           6     18    54 (can you see a pattern yet? OK we'll go to the second line)

               ↔    ↔

               12    36 ( Yes they are multiples of 3,this suggests the rule

involves 3 ^{to the power} of something)

 

Therefore;

u_n = 3ⁿ (this is the start of our formula)

Lets put 3ⁿ into the equation and see what else we need!

3¹ = 3 but we need 8 so we must need to add 5

So now we'll try the formula 3ⁿ + 5

3¹ = 3 + 5 = 8

3² = 9 + 5 = 14

3³ = 27 + 5 = 32

3⁴ = 81 + 5 = 86

if you check back to the above table the formula is correct

u_n = 3n + 5

 

 

 

(iii) Now we'll do one with fractions ¹/2, ¹/7, ¹/12, ¹/17

 

Lets put these values into a table

n     1      2      3       4

u_n  ¹/2   ¹/7   ¹/12   ¹/17 (notice all of them have the numerator 1)

           ↔     ↔      ↔ ( the pattern in the denominator is going up by 5)

            5       5       5

This is a simple pattern!

So u_n = 5n + or - ?

5 x 1 = 5 but we need 2 so we'll take away 3

Our formula now looks like this; 5n - 3 (to get the bottom number in the fractions)

5 x 1 = 5 - 3 = 2

5 x 2 = 10 - 3 = 7

5 x 3 = 15 - 3 = 12

5 x 4 = 20 - 3 = 17

and on the top we have 1

So our formula now looks like this u_n = ¹/ 5n - 3

u_n = ¹/ 5n - 3