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(n+1)(n-1) Indices

Hi, this may be sixth form, not sure.

How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

So, I am supposed to show that Un−1Un+1−Un2=12n−1\displaystyle U_{n-1}U_{n+1}-U^2_n=12^{n-1} , for n=1,2,3...\displaystyle n=1,2,3...

Un=3n+4n\displaystyle U_n=3^n+4^n

Un−1Un+1=(3n−1+4n−1)(3n+1+4n+1)=32n+3n−14n+1+4n−13n+1+42n\displaystyle U_{n-1}U_{n+1}=(3^{n-1}+4^{n-1})(3^{n+1}+4^{n+1}) \\ \\ =3^{2n}+3^{n-1}4^{n+1}+4^{n-1}3^{n+1}+4^{2n}

fine


Un2=(3n+4n)2=32n+2(3n4n)+42n\displaystyle U_n^2=(3^n+4^n)^2=3^{2n}+2(3^n4^n)+4^{2n}

fine


Now, what I don't get is how this line becomes the next:s-smilie:

3n−14n+1+4n−13n+1−2(3n4n)\displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


=3n−14n−1(42+32−2×3×4)\displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

How does that work?
They factorised it using the laws of indices.
Reply 2
Avatar for anil10100
anil10100
Original post by SubAtomic
Hi, this may be sixth form, not sure.

How does this work, cannot quite fathom it, no help in book so am left in ambiguity.

So, I am supposed to show that Un−1Un+1−Un2=12n−1\displaystyle U_{n-1}U_{n+1}-U^2_n=12^{n-1} , for n=1,2,3...\displaystyle n=1,2,3...

Un=3n+4n\displaystyle U_n=3^n+4^n

Un−1Un+1=(3n−1+4n−1)(3n+1+4n+1)=32n+3n−14n+1+4n−13n+1+42n\displaystyle U_{n-1}U_{n+1}=(3^{n-1}+4^{n-1})(3^{n+1}+4^{n+1}) \\ \\ =3^{2n}+3^{n-1}4^{n+1}+4^{n-1}3^{n+1}+4^{2n}

fine


Un2=(3n+4n)2=32n+2(3n4n)+42n\displaystyle U_n^2=(3^n+4^n)^2=3^{2n}+2(3^n4^n)+4^{2n}

fine


Now, what I don't get is how this line becomes the next:s-smilie:

3n−14n+1+4n−13n+1−2(3n4n)\displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


=3n−14n−1(42+32−2×3×4)\displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

How does that work?


Just notice that 3n−14n+1=3n−14n−1+2=3n−14n−142 3^{n-1}4^{n+1}=3^{n-1}4^{n-1+2}=3^{n-1}4^{n-1}4^2 , similar thing for the others.
Reply 3
Avatar for SubAtomic
SubAtomic
OP
11
Original post by anil10100
Just notice that 3n−14n+1=3n−14n−1+2=3n−14n−142 3^{n-1}4^{n+1}=3^{n-1}4^{n-1+2}=3^{n-1}4^{n-1}4^2 , similar thing for the others.


Nice, will give it a whirl and see where I end up.
Reply 4
Avatar for SubAtomic
SubAtomic
OP
11
So are these equivalent then

an−1bn+1≡(ab)n−1b2\displaystyle a^{n-1}b^{n+1} \equiv (ab)^{n-1}b^2

Seems to be according to the calculator.


Back to the question, so

3n−14n−1+2+4n−13n−1+2−2(3n4n)\displaystyle 3^{n-1}4^{n-1+2}+4^{n-1}3^{n-1+2}-2(3^n4^n)


=3n−14n−142+3n−14n−132−2(3n4n)\displaystyle = 3^{n-1}4^{n-1}4^2+3^{n-1}4^{n-1}3^2-2(3^n4^n)

Is this right? What now? How do I get rid of those n in the bracket so it looks like the last bit of my OP?
Reply 5
Avatar for raheem94
raheem94
13
Original post by SubAtomic
So are these equivalent then

an−1bn+1≡(ab)n−1b2\displaystyle a^{n-1}b^{n+1} \equiv (ab)^{n-1}b^2

Seems to be according to the calculator.


In general, ax×ay=ax+y a^x \times a^y = a^{x+y}

an−1bn+1=an−1bn−1+2=an−1b(n−1)+2=an−1b(n−1)×b2=(ab)n−1b2 \displaystyle a^{n-1}b^{n+1} = a^{n-1} b^{ n - 1 + 2 } = a^{n-1} b^{ (n - 1) + 2 } = a^{n-1} b^{ (n - 1) } \times b^2 = (ab)^{n-1} b^2
Reply 6
Avatar for SubAtomic
SubAtomic
OP
11
Original post by raheem94
In general, ax×ay=ax+y a^x \times a^y = a^{x+y}

an−1bn+1=an−1bn−1+2=an−1b(n−1)+2=an−1b(n−1)×b2=(ab)n−1b2 \displaystyle a^{n-1}b^{n+1} = a^{n-1} b^{ n - 1 + 2 } = a^{n-1} b^{ (n - 1) + 2 } = a^{n-1} b^{ (n - 1) } \times b^2 = (ab)^{n-1} b^2


Yeah that was my own little spin on things, not too sure on the second bit of that post.
(edited 11 years ago)
Reply 7
Avatar for raheem94
raheem94
13
Original post by SubAtomic
Yeah that was my own little spin on things, not too sure on the second bit of that post.


Spoiler

(edited 11 years ago)
Reply 8
Avatar for aznkid66
aznkid66
1
Original post by SubAtomic

Now, what I don't get is how this line becomes the next:s-smilie:

3n−14n+1+4n−13n+1−2(3n4n)\displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)


=3n−14n−1(42+32−2×3×4)\displaystyle = 3^{n-1}4^{n-1}(4^2+3^2-2 \times 3 \times 4)

How does that work?


3n−14n+1+4n−13n+1−2(3n4n)\displaystyle 3^{n-1}4^{n+1}+4^{n-1}3^{n+1}-2(3^n4^n)

For easier comprehension, let's break it down by isolating the 3n−14n−13^{n-1}4^{n-1} in each individual term.



First term:

3n−14n+1\displaystyle 3^{n-1}4^{n+1}

Spoiler



(3n−14n−1)42(3^{n-1}4^{n-1})4^2



Second term:

4n−13n+14^{n-1}3^{n+1}

Spoiler



(3n−14n−1)32(3^{n-1}4^{n-1})3^2



Third term:

−2(3n4n)-2(3^n4^n)

Spoiler



−(3n−14n−1)(2)(3)(4)-(3^{n-1}4^{n-1})(2)(3)(4)



All together now:

(3n−14n−1)42+(3n−14n−1)32−(3n−14n−1)(2)(3)(4)(3^{n-1}4^{n-1})4^2+(3^{n-1}4^{n-1})3^2-(3^{n-1}4^{n-1})(2)(3)(4)

Factorize out 3^(n-1)*4^(n-1):

(3n−14n−1)(42+32−(2)(3)(4))(3^{n-1}4^{n-1})(4^2+3^2-(2)(3)(4))
(edited 11 years ago)
Reply 9
Avatar for SubAtomic
SubAtomic
OP
11
Original post by aznkid66
...


Nice explanation again, I thought of subbing the n=(n-1)+1 into the−2(3n4n) -2(3^n4^n) but thought it a bit of an odd way of doing things for some reason:rolleyes: you took it that extra step, thanks. You are missing the closing brackets in your third spoiler but I get ya:smile:


Original post by raheem94
...


Nice one mate, both your explanations have helped.


Must take quite a while to be able to rep people again:frown:
(edited 11 years ago)
Reply 10
Avatar for SubAtomic
SubAtomic
OP
11
So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

This first

Un−4Un+8−Un2\displaystyle U_{n-4}U_{n+8}-U_n^2 for n=1,2,3...\displaystyle n=1,2,3...


Un=5n+6n\displaystyle U_n=5^n+6^n


Un−4Un+8=(5n−4+6n−4)(5n+8+6n+8)=52n+4+5n−46n+8+6n−45n+8+62n+4\displaystyle U_{n-4}U_{n+8}=(5^{n-4}+6^{n-4})(5^{n+8}+6^{n+8}) \\ \\ =5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}

Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?

Will carry on as per anyway,

Un2=(5n+6n)2=52n+2(5n6n)+62n\displaystyle U_n^2=(5^n+6^n)^2=5^{2n}+2(5^n6^n)+6^{2n}

So


Un−4Un+8−Un2=52n+4+5n−46n+8+6n−45n+8+62n+4−(52n+2(5n6n)+62n)\displaystyle U_{n-4}U_{n+8}-U_n^2=5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}-(5^{2n}+2(5^n6^n)+6^{2n})


=54+5n−46n−4612+6n−45n−4512+64−2(5n−46n−4)5464\displaystyle = 5^4+5^{n-4}6^{n-4}6^{12}+6^{n-4}5^{n-4}5^{12}+6^4-2(5^{n-4}6^{n-4})5^46^4


=5n−46n−4(612+512−2×54×64)+54+64=30n−4(612+512−2×54×64)+54+64\displaystyle = 5^{n-4}6^{n-4}(6^{12}+5^{12}-2 \times 5^4 \times 6^4)+5^4+6^4 \\ \\ = 30^{n-4}(6^{12}+5^{12}-2 \times 5^4 \times 6^4)+5^4+6^4


=30n(612+512−2×54×64)304+54+64\displaystyle=\dfrac{30^n(6^{12}+5^{12}-2 \times 5^4 \times 6^4)}{30^4}+5^4+6^4

Is this right? If it is can I simplify further?
(edited 11 years ago)
Reply 11
Avatar for raheem94
raheem94
13
Original post by SubAtomic
So let me see if I have got this right by trying another, tried one earlier and it went horribly wrong, I think.

This first

Un−4Un+8−Un2\displaystyle U_{n-4}U_{n+8}-U_n^2 for n=1,2,3...\displaystyle n=1,2,3...


Un=5n+6n\displaystyle U_n=5^n+6^n


Un−4Un+8=(5n−4+6n−4)(5n+8+6n+8)=52n+4+5n−46n+8+6n−45n+8+62n+4\displaystyle U_{n-4}U_{n+8}=(5^{n-4}+6^{n-4})(5^{n+8}+6^{n+8}) \\ \\ =5^{2n+4}+5^{n-4}6^{n+8}+6^{n-4}5^{n+8}+6^{2n+4}

Can I do something with 2^2 in the indices here? Seen as 4 and 8 are 2^2 and 2^3. Or would I be over-complicating things?


52n+4=52n×54=(52)n×54=25n×625=625(25)n 5^{ 2n + 4} = 5^{2n} \times 5^4 = (5^2)^n \times 5^4 = 25^n \times 625 = 625 (25)^n


5n−46n+8+6n−45n+8=5n5−46n68+6n6−45n58=5n6n(5−468+6−458)=(5×6)n(5−468+6−458)=30n(5−468+6−458) 5^{n-4}6^{n+8}+6^{n-4}5^{n+8} = 5^n 5^{-4} 6^n 6^8 + 6^n 6^{-4} 5^n 5^8 = 5^n 6^n ( 5^{-4} 6^8 + 6^{-4} 5^8 ) \\ = ( 5 \times 6)^n ( 5^{-4} 6^8 + 6^{-4} 5^8 ) = 30^n (5^{-4} 6^8 + 6^{-4} 5^8 )


62n+4=62n×64=(62)n×64=1296(36)n 6^{2n+4} = 6^{2n} \times 6^4 = (6^2)^n \times 6^4 = 1296(36)^n
(edited 11 years ago)
Reply 12
Avatar for SubAtomic
SubAtomic
OP
11
Original post by raheem94
...


Yep I understand this, was meaning something like 52n+4≡521n+22≡5(n+2)2\displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2}

6n−45n+8≡6n−(22)5n+(23)\displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)}

Just trying stuff nothing major, seeing if a neater way.
(edited 11 years ago)
Reply 13
Avatar for raheem94
raheem94
13
Original post by SubAtomic
Yep I understand this, was meaning something like 52n+4≡521n+22≡5(n+2)2\displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2}

6n−45n+8≡6n−(22)5n+(23)≡30(n−4)+(n+8)\displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)} \equiv 30^{(n-4)+(n+8)}

Just trying stuff nothing major, seeing if a neater way.


52n+4≡521n+22≡5(n+2)2≡(52)n+2≡25n+2≡25n×252=625(25)n\displaystyle 5^{2n+4} \equiv 5^{2^1n+2^2} \equiv 5^{(n+2)2} \equiv (5^2)^{n+2} \equiv 25^{n+2} \equiv 25^n \times 25^2 = 625(25)^n
Reply 14
Avatar for raheem94
raheem94
13
Original post by SubAtomic

6n−45n+8≡6n−(22)5n+(23)≡30(n−4)+(n+8)\displaystyle 6^{n-4}5^{n+8} \equiv 6^{n-(2^2)}5^{n+(2^3)} \equiv 30^{(n-4)+(n+8)}


This is wrong.

6n−45n+8≢30(n−4)+(n+8) 6^{n-4}5^{n+8} \not\equiv 30^{(n-4)+(n+8)}

If you want to check it, sub in a number.

Let n=1,6n−45n+8=6−359=9042.2453730(n−4)+(n+8)=30−3+9=306=729000000 \displaystyle \text{Let } n = 1, \\ 6^{n-4}5^{n+8} = 6^{-3}5^9 = 9042.24537 \\ 30^{(n-4)+(n+8)} = 30^{-3 + 9 } = 30^6 =729000000
Reply 15
Avatar for SubAtomic
SubAtomic
OP
11
Original post by raheem94
..


Edited, experiments and all.
Reply 16
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SubAtomic
OP
11
Original post by raheem94
...


The main thing I was wondering about is how the indices have common factors and if they can be simplified in a certain fashion, not actually quite sure what I was thinking but it was something lol.

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