# 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 **2**^{n }(2 to the power of ^{"n"})

This can be written as **u _{n} = 2^{n}**

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} = 2^{n}**

When** ^{"n"} is 1**...

^{}2

^{1}(2 to the power of

^{1}) =

**2**

when ** ^{"n" }is 2**... 2

^{2}(2 to the power of

^{2}) =

**4**

when ** ^{"n" }is 3**... 2

^{3}(2 to the power of 3) =

**8**

when ** ^{"n" }is 4**... 2

^{4}(2 to the power of 4) =

**16**

when ** ^{"n" }is 5**... 2

^{5}(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 }= 2

^{0 }which = 1 + 1(n) =

**2**

When ** ^{"n"} is 2**... 2

^{2 - 1 }=

^{}2

^{1 }which = 2 + 2(n) =

**4**

When ** ^{"n"} is 3**... 2

^{3 - 1 }= 2

^{2}which = 4 + 3(n) =

**7**

When ** ^{"n"} is 4**... 2

^{4 - 1 }= 2

^{3 }which = 8 + 4(n) =

**12**

When ** ^{"n"} is 5**... 2

^{5 - 1 }= 2

^{4}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}= 4

^{0 }which =

**1**

When ** ^{"n"} is 3**... 4

^{3 - 2 }= 4

^{1 }which =

**4**

When ** ^{"n"} is 4**... 4

^{4 - 2 }= 4

^{2 }which =

**16**

When ** ^{"n"} is 5**... 4

^{5 - 2 }= 4

^{3 }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

**n**

^{2}^{}

**u _{n} = ?**

**u _{n} = n^{2}**

**If we square the 1 ^{2} that = 1, but we need 3 so we'll multiply by 3 ( 1 x 3 = 3)**

**Now to square the 2 ^{2 }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 n^{2}**

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

**u _{n} = 3 n^{2}**

**u _{n} = 3 x 3^{2} = 3 x 9 = 27**

**u _{n} = 3 x 4^{2 }= 3 x 16 = 48**

**u _{n} = 3 x 5^{2 }= 3 x 25 = 75**

**so the formula is correct **

**u _{n} = 3 n^{2}**

^{}

^{}

^{}

**( 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} = 3^{n} - 1**

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

^{}

**3 ^{1 }= 3** ( but if you look at the table we only need

**2,**so we'll take

**1**away)

Our formula will now be **3 ^{n} - 1**

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

**3 ^{2}**

^{}

**= 9 - 1 = 8**

**3 ^{3} = 27 - 1 = 26**

**3 ^{4 }= 81 - 1 = 80**

**3 ^{5} = 243 - 1 = 242**

**so the formula is correct **

**u _{n} = 3^{n} - 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} = 2^{n}**

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

**2 ^{1 }= 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} = 2^{n }+ 1**

**Lets SEE!**

**2 ^{1 }= 2 + 1 = 3 (**

__ok so far but look carefully__

**)**

**Next one......**

**2 ^{2} = 4 + 1 = 5 (**

__this isn't going to work as we need__

**6**, see above table**)**

**So u _{n} = 2^{n }+ 1 is wrong!**

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

Our** formula** now is **u _{n} = 2^{n }+ n**

**Therefore;**

**2 ^{1 }= 2 + 1 = 3**

**2 ^{2 }= 4 + 2 = 6**

**2 ^{3} = 8 + 3 = 11**

**2 ^{4 }= 16 + 4 = 20**

**2 ^{5 }= 32 + 5 = 37 **

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

**u _{n} = 2^{n }+ 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 }= **

^{}

**2**

^{n }

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

Therefore; **u _{n }= **

^{}

**2**

^{n - 1 }

**2 ^{1 - 1 }= n^{0} = 2**

^{0 }which equals

**1**

**2 ^{2 - 1} = n^{1 }= 2^{1 }**which equals

**2**

**2 ^{3 - 1 }= n^{2} = 2^{2 }**which equals

**4**

**2 ^{4 - 1 }= n^{3} = 2^{3 }**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} = 3^{n }(this is the start of our formula)**

**Lets put 3 ^{n} into the equation and see what else we need!**

**3 ^{1} = 3** but we need

**8**so we must need to

**add 5**

**So now we'll try the formula 3 ^{n} + 5**

**3 ^{1 }= 3 + 5 = 8**

**3 ^{2} = 9 + 5 = 14**

**3 ^{3} = 27 + 5 = 32**

**3 ^{4 }= 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**

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