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

Updated on October 4, 2010

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

Complete the tables below given that the nth term is:

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

This can be written as un = 2n

So lets start to substitute

n     1     2     3     4     5

un    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.)

un = 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 nth term is 2n - 1 + n (2 to the power of "n -1" + n)

This can be written as un = 2n - 1 + n

So lets start to substitute

n     1     2     3     4      5

un    2     4     7     12    21

un = 2n - 1 + n

When "n" is 1... 21 - 1 = 20 which = 1 + 1(n) = 2

When "n" is 2... 22 - 1 =21 which = 2 + 2(n) = 4

When "n" is 3... 23 - 1 = 22 which = 4 + 3(n) = 7

When "n" is 4... 24 - 1 = 23 which = 8 + 4(n) =12

When "n" is 5... 25 - 1 = 24 which =16 + 5(n) =21

♦ Note that "n0" will always be 1

(iii) Now to complete the table where the nth term is 4n - 2(4 to the power of "n - 2")

This can be written as un = 4n - 2

So lets start to substitute

n     1      2     3     4     5

un 0.25   1     4    16   64

un = 4n - 2

When "n" is 1... 41 - 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... 42 - 2 = 40 which = 1

When "n" is 3... 43 - 2 = 41 which = 4

When "n" is 4... 44 - 2 = 42 which = 16

When "n" is 5... 45 - 2 = 43 which = 64

Now to find the formula for un in the following tables

( i )

n     1     2     3     4     5

un    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 un , above since the second line has a constant number this suggests we are looking for n2

un = ?

un = 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

un = 3 n2

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

un = 3 n2

un = 3 x 32 = 3 x 9 = 27

un = 3 x 42 = 3 x 16 = 48

un = 3 x 52 = 3 x 25 = 75

so the formula is correct

un = 3 n2

( ii ) Example

n     1     2     3     4      5

un     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)

un = ?

un = 3n - 1

So following our rule which suggests 3to 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

un = 3n - 1

( iii ) Next Example

n     1     2     3     4     5

un     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 2to the power of something)

un = ?

un = 2n

So following our rule which suggests 2to 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 un = 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 un = 2n + 1 is wrong!

Our formula now is un = 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

un = 2n + n

Find the nth 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

un     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.)

un = ?

un = 2n

This time the formula involves subtraction of the power

Therefore; un = 2n - 1

21 - 1 = n0 = 20 which equals 1

22 - 1 = n1 = 21 which equals 2

23 - 1 = n2 = 22 which equals 4

24 - 1 = n3 = 23 which equals 8

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

un = 2n - 1

(ii) 8, 14, 32, 86

Lets put these values into a table

n     1     2     3     4

un     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;

un = 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

un = 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

u1/2   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 un = 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 un = 1/ 5n - 3

un = 1/ 5n - 3

7

0

10

16

26

working