-
Best of luck to the class of 2024 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
4u Mathematics Marathon V 1.0 (2 Viewers)
- Thread starter who_loves_maths
- Start date
onebytwo
Recession '08
sorry, just did a late edit
haque
Member
- Joined
- Sep 2, 2006
- Messages
- 426
- Gender
- Male
- HSC
- 2006
first we find a^3+b^3+c^3 by subsituting into the equations and finding a^2+b^2+c^2 through a sort of recursion formula-we apply that to the a^4+b^4+c^4 case by multiplying through each equation by a,b,c and end up getting 386
Next question
Show that (sigma sign infinity to 1 1/k(k+1)(k+2).....(k+p)) converges to 1/p*p!
Next question
Show that (sigma sign infinity to 1 1/k(k+1)(k+2).....(k+p)) converges to 1/p*p!
Last edited:
onebytwo
Recession '08
wait..wait..
im a bit slow
Lotto, can you explain your first line.....it doesnt make sense to me...
im a bit slow
Lotto, can you explain your first line.....it doesnt make sense to me...
onebytwo
Recession '08
so...
your saying cosec2x= 1-cosec2x
EDIT: thought somthing didnt make sense
ok moving on..
EDIT: sorry about this
on the third line on the RHS, shouldnt you have tan2x +cot2
your saying cosec2x= 1-cosec2x
EDIT: thought somthing didnt make sense
ok moving on..
EDIT: sorry about this
on the third line on the RHS, shouldnt you have tan2x +cot2
Last edited:
Ans to haque's partial fraction bashfest:
the q is slightly wrong imo, it is sigma[k=1 to k=infinity][1/k(k+1)...(k+p)]
Split this fraction into 2 parts:
1/k(k+1)...(k+p) = 1/p[1/k(k+1)...(k+p-1) - 1/(k+1)(k+2)...(k+p)]
when k=1, 1/1.2.3...(p+1) = 1/p[[1/1.2.3...p]-[1/2.3.4...(p+1)]]
when k=2, 1/2.3.4...(p+2) = 1/p[[1/2.3.4...(p+1)]-[1/3.4.5...(p+2)]]
when k=3, 1/3.4.5...(p+3)=1/p[[1/3.4.5...(p+2)]-[1/4.5.6...(p+3)]]
...
when k=n, 1/n(n+1)(n+2)...(n+p)=1/p[[1/n(n+1)...(n+p-1)]-[1/(n+1)(n+2)...(n+p)]
This is a telescoping sum, when u add the above statements, every expression on the RHS cancels except the first and last
ie
sigma[k=1 to k=infinity][1/k(k+1)...(k+p)]=lim[n->infinity]1/p[[1/p!]-[1/(n+1)(n+2)...(n+p)]]
=1/p*p!
as required
nice q haque, where is it from? i havent seen many examples on telescoping sums in any 4u textbooks
the q is slightly wrong imo, it is sigma[k=1 to k=infinity][1/k(k+1)...(k+p)]
Split this fraction into 2 parts:
1/k(k+1)...(k+p) = 1/p[1/k(k+1)...(k+p-1) - 1/(k+1)(k+2)...(k+p)]
when k=1, 1/1.2.3...(p+1) = 1/p[[1/1.2.3...p]-[1/2.3.4...(p+1)]]
when k=2, 1/2.3.4...(p+2) = 1/p[[1/2.3.4...(p+1)]-[1/3.4.5...(p+2)]]
when k=3, 1/3.4.5...(p+3)=1/p[[1/3.4.5...(p+2)]-[1/4.5.6...(p+3)]]
...
when k=n, 1/n(n+1)(n+2)...(n+p)=1/p[[1/n(n+1)...(n+p-1)]-[1/(n+1)(n+2)...(n+p)]
This is a telescoping sum, when u add the above statements, every expression on the RHS cancels except the first and last
ie
sigma[k=1 to k=infinity][1/k(k+1)...(k+p)]=lim[n->infinity]1/p[[1/p!]-[1/(n+1)(n+2)...(n+p)]]
=1/p*p!
as required
nice q haque, where is it from? i havent seen many examples on telescoping sums in any 4u textbooks
Let I<sub>n</sub>=∫<sub>0</sub><sup>1</sup>(e^t)(t^n)dt
(a) Prove by induction that
I<sub>n</sub>=[(-1)^(n+1)]n!+e[sigma i=0 to i=n][(-1)^i]n!/(n-i)!
(b)Show that 1/(n+1)<=I<sub>n</sub><=∫<sub>0</sub><sup>1</sup>e(t^n)dt<=e/n
(c) Hence show that e is irrational
(a) Prove by induction that
I<sub>n</sub>=[(-1)^(n+1)]n!+e[sigma i=0 to i=n][(-1)^i]n!/(n-i)!
(b)Show that 1/(n+1)<=I<sub>n</sub><=∫<sub>0</sub><sup>1</sup>e(t^n)dt<=e/n
(c) Hence show that e is irrational
Last edited:
haque
Member
- Joined
- Sep 2, 2006
- Messages
- 426
- Gender
- Male
- HSC
- 2006
Yea it's just the bit on -1 to the power of n on n! i was wondering if there was anything else multiplying it like i thought it might be (-1)^n+1 *n^2/(n!) cos in the induction everything fits in except that bit-maybe i'm doing something wrong i'll check again-right now i gotta do some modern history-cya yip and good luck.
onebytwo
Recession '08
*bump*
a couple easy non-topical questions to warm everyone up
- the letters abc....zABC...Z are arranged in a line at random. what is the probability that the lower case letters and the upper case letters are in alphabetical order?
- if a and b are positive integers, prove that (1/a) + (1/b) >= (4/(a+b))
a couple easy non-topical questions to warm everyone up
- the letters abc....zABC...Z are arranged in a line at random. what is the probability that the lower case letters and the upper case letters are in alphabetical order?
- if a and b are positive integers, prove that (1/a) + (1/b) >= (4/(a+b))
For the first one:
Arranging lower case letters in alphabetical order= 1 way
Arranging upper case letters in alphabetical order= 1 way
Arranging these two sets amongst themselves = 2 ways
.: Total = 2 ways
Arranging of letters without restriction = 52!
.(upper and lower in order)= 2/52!
Arranging lower case letters in alphabetical order= 1 way
Arranging upper case letters in alphabetical order= 1 way
Arranging these two sets amongst themselves = 2 ways
.: Total = 2 ways
Arranging of letters without restriction = 52!
.
(upper and lower in order)= 2/52!
Last edited:
For the second one:
(a-b)2>=0
(a-b)2 + 4ab>=4ab
(a+b)2>=4ab
a+b>=(4ab)/(a+b)
.: 1/b + 1/a>=4/(a+b)
New Question:
A particle of weight 4 newton, intiially at rest, falls vertically in a resisting medium with a retarding force of kv newton, where k is a constant and v m/s is the velocity of the particle at any time t seconds after release.
a) Evaluate k if the particle eventually attains a constant speed of 20m/s
b) Find the velocity of the particle at any time t
(a-b)2>=0
(a-b)2 + 4ab>=4ab
(a+b)2>=4ab
a+b>=(4ab)/(a+b)
.: 1/b + 1/a>=4/(a+b)
New Question:
A particle of weight 4 newton, intiially at rest, falls vertically in a resisting medium with a retarding force of kv newton, where k is a constant and v m/s is the velocity of the particle at any time t seconds after release.
a) Evaluate k if the particle eventually attains a constant speed of 20m/s
b) Find the velocity of the particle at any time t
Last edited:
onebytwo
Recession '08
i wish i had the answers for these, anywayzeek said:For the first one:
Arranging lower case letters in alphabetical order= 1 way
Arranging upper case letters in alphabetical order= 1 way
Arranging these two sets amongst themselves = 2 ways
.: Total = 2 ways
Arranging of letters without restriction = 52!
.
i got a different answer.....1/(26!)2