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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!

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

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

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    • t.elia profile imageAUTHOR

      t.elia 

      5 years ago from Northern Ireland

    • t.elia profile imageAUTHOR

      t.elia 

      8 years ago from Northern Ireland

      For further help follow the link in the hub "Let me solve your maths problems". Assistance can be found for

      Bodmas

      HCF

      LCM

      Prime and Composite numbers

      Square Root

      Cube Root

      Place Values

      Expanded Notation

      Rounding off Numbers

      Writing numbers as a product of their Prime Numbers

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