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2 questions: poly, then other (1 Viewer)
- Thread starter chousta
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student.hsc
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- Nov 2, 2006
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- HSC
- 2007
Ok;
1) (not gonna show working)
Method: expansion, equating like terms, simultaneous eqn
Answer: A= 28/15
B= 17/15
C= -22/15
Thats wat i got newaz...
2)hmm... I cant read the subscripts but this certainly looks like induction
starting with
n=1 works (given)
n=2 works (also given)
Assume true for n=k
Prove for n=k+1
This should do it, but I cant read it...
1) (not gonna show working)
Method: expansion, equating like terms, simultaneous eqn
Answer: A= 28/15
B= 17/15
C= -22/15
Thats wat i got newaz...
2)hmm... I cant read the subscripts but this certainly looks like induction
starting with
n=1 works (given)
n=2 works (also given)
Assume true for n=k
Prove for n=k+1
This should do it, but I cant read it...
P
pLuvia
Guest
1. If you want to check the answers, just simiplify the answer you obtained and see if it equals the RHS
2.
Can't read it, but just follow the normal procedures
2.
Can't read it, but just follow the normal procedures
If you assume that uk = (k+3).2k, and uk-1 = (k+2).2k-1 then, given un = 4un-1 - 4un-2, we obtain:
uk+1 = 4uk - 4uk-1
= 4.(k+3).2k - 4.(k+2).2k-1
= (2k+6).2k+1 - (k+2).2k+1
= [(k+1] + 3).2k+1 , as required (or so it would seem)
(Edit, for the sake of completion: You can easily show that u3 and u4 satisfy the condition they require - just shove numbers into the formulae - which is why you can then make the assumption for both uk AND uk-1)
uk+1 = 4uk - 4uk-1
= 4.(k+3).2k - 4.(k+2).2k-1
= (2k+6).2k+1 - (k+2).2k+1
= [(k+1] + 3).2k+1 , as required (or so it would seem)
(Edit, for the sake of completion: You can easily show that u3 and u4 satisfy the condition they require - just shove numbers into the formulae - which is why you can then make the assumption for both uk AND uk-1)
Last edited:
1. A=2, B=1, C=-2 by cross multiplying, then substituting x=4 which gets the A value, after which B and C come easily
2. K-funk's method is the straightforward HSC method that should be used in exams. If you want to, you can do it without induction (only do this in an exam if you have a ton of spare time lol)
u<sub>n</sub> = 4u<sub>n-1</sub> - 4u<sub>n-2
</sub> u<sub>n</sub>-2u<sub>n-1</sub>= 2u<sub>n-1</sub> - 4u<sub>n-2</sub>=2(u<sub>n-1</sub>-2u<sub>n-2</sub>)
u<sub>n-1</sub>-2u<sub>n-2</sub>=2(u<sub>n-2</sub>-2u<sub>n-3</sub>)
...
u<sub>3</sub>-2u<sub>2</sub>=2(20-16)=2<sup>3</sup>
From above, by repeatedly substituting the lhs into the rhs of the equation above it,
u<sub>n</sub>-2u<sub>n-1</sub>=2<sup>n</sup>
Dividing both sides by 2<sup>k-1 </sup>gives
(u<sub>n</sub>/2<sup>n</sup>)-(u<sub>n-1</sub>/2<sup>n-1</sup>)=1
(u<sub>n-1</sub>/2<sup>n-1</sup>)-(u<sub>n-2</sub>/2<sup>n-2</sup>)=1
...
(u<sub>2</sub>/2<sup>2</sup>)-(u<sub>1</sub>/2)=1
Adding the above equations,
(u<sub>n</sub>/2<sup>n</sup>)-(u<sub>1</sub>/2)=n-2
u<sub>n</sub>/2<sup>n</sup>=n+3
u<sub>n</sub>=2<sup>n</sup>(n+3)
<sup>
</sup><sub></sub><sub></sub><sub></sub>
2. K-funk's method is the straightforward HSC method that should be used in exams. If you want to, you can do it without induction (only do this in an exam if you have a ton of spare time lol)
u<sub>n</sub> = 4u<sub>n-1</sub> - 4u<sub>n-2
</sub> u<sub>n</sub>-2u<sub>n-1</sub>= 2u<sub>n-1</sub> - 4u<sub>n-2</sub>=2(u<sub>n-1</sub>-2u<sub>n-2</sub>)
u<sub>n-1</sub>-2u<sub>n-2</sub>=2(u<sub>n-2</sub>-2u<sub>n-3</sub>)
...
u<sub>3</sub>-2u<sub>2</sub>=2(20-16)=2<sup>3</sup>
From above, by repeatedly substituting the lhs into the rhs of the equation above it,
u<sub>n</sub>-2u<sub>n-1</sub>=2<sup>n</sup>
Dividing both sides by 2<sup>k-1 </sup>gives
(u<sub>n</sub>/2<sup>n</sup>)-(u<sub>n-1</sub>/2<sup>n-1</sup>)=1
(u<sub>n-1</sub>/2<sup>n-1</sup>)-(u<sub>n-2</sub>/2<sup>n-2</sup>)=1
...
(u<sub>2</sub>/2<sup>2</sup>)-(u<sub>1</sub>/2)=1
Adding the above equations,
(u<sub>n</sub>/2<sup>n</sup>)-(u<sub>1</sub>/2)=n-2
u<sub>n</sub>/2<sup>n</sup>=n+3
u<sub>n</sub>=2<sup>n</sup>(n+3)
<sup>
</sup><sub></sub><sub></sub><sub></sub>