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    (Original post by Occams Chainsaw)
    Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do...
    We just started FP1 so he was trying to play with it!

    (2n - 1) = n^2

n = 1: 1=1 check!



(2k - 1) = k^2 



k+1: 2(k + 1) - 1 = (k + 1)^2

2(k + 1) - 1 = 2(k + 1) - 1

k^2 + (2(k + 1) - 1) = (k + 1)^2 

2(k + 1) - 1 = (k + 1)^2
    Not even sure if that makes sense right now. Brain = fried.
    I know it's not induction. That's why I was questioning it.

    I honestly cannot see what you are trying to do....That top line isn't an identity.
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    (Original post by bananarama2)
    I know it's not induction. That's why I was questioning it.

    I honestly cannot see what you are trying to do....That top line isn't an identity.
    I think I'll come fix it in the morning after some sleep and save further embarrassment.

    I think I just proved that  1 + 3 + 5 + 7 + 9... + (2n-1)=n^2 , didn't I?
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    (Original post by Occams Chainsaw)
    I think I'll come fix it in the morning after some sleep and save further embarrassment.

    I think I just proved that  1 + 3 + 5 + 7 + 9... + (2n-1)=n^2 , didn't I?
    Aha I think that's wise.

    It's not what is written, but I think it may be have been latexed badly
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    (Original post by bananarama2)
    Aha I think that's wise.

    It's not what is written, but I think it may be have been latexed badly
    I've not used latex before :sexface:
    I guess that ones been done a few times, right? .... :sexface:

    Anyway, I hope to redeem myself after some kip!... :sexface: It's exactly difficult lol. I'm just plugging in k+1!
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    (Original post by Occams Chainsaw)
    I've not used latex before :sexface:
    I guess that ones been done a few times, right? .... :sexface:

    Anyway, I hope to redeem myself after some kip!... :sexface: It's exactly difficult lol. I'm just plugging in k+1!
    Spoiler-How it should be done


    Statement :  \displaystyle \sum_{i=1}^n 2i-1=n^2

    Proof: When n=2,  (2-1)+(4-1)=4=2^2

    So true for n=1.

    Assume true for n=k.

    For n= k+1:

     \displaystyle \sum_{i=1}^{k+1} 2i-1

     \displaystyle = \sum_{i=1}^{k} (2i-1) + 2(k+1)-1

     \displaystyle = k^2 + 2(k+1)-1

     \displaystyle = (k+1)^2

    True for n=k+1, therefore true for postive integers.

    Which is sort of what you wrote without the sigmas, which makes me think you did know what you were doing



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    Any of you made up these questions yourself?

    Posted from TSR Mobile
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    (Original post by bananarama2)
    Spoiler-How it should be done


    Statement :  \displaystyle \sum_{i=1}^n 2i-1=n^2

    Proof: When n=2,  (2-1)+(4-1)=4=2^2

    So true for n=1.

    Assume true for n=k.

    For n= k+1:

     \displaystyle \sum_{i=1}^{k+1} 2i-1

     \displaystyle = \sum_{i=1}^{k} (2i-1) + 2(k+1)-1

     \displaystyle = k^2 + 2(k+1)-1

     \displaystyle = (k+1)^2

    True for n=k+1, therefore true for postive integers.

    Which is sort of what you wrote without the sigmas, which makes me think you did know what you were doing



    Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?!
    What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!

    A nice little pattern I noticed is this:
    if I have an n*n box - or a perfect square
    xxxx
    xxxx
    xxxx
    xxxx

    in this case where n = 4, x = 16

    where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)

    abcd
    abcc
    abbb
    aaaa

    So we have produced 7a + 5b + 3b + 1a = 16x
    or, 16 (a square number) = 7 + 5 + 3 + 1...

    Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol
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    (Original post by jack.hadamard)
    Problem 345 **/***

    Evaluate \displaystyle \lim_{x \to 0} \frac{\displaystyle \int_{0}^{\tan x} \tan\sin t\ dt}{\displaystyle \int_{0}^{\sin x} \ln(1 + \sin t)\ dt}.
    An alternative solution to the one given:

    Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have

    \tan(\sin(t))=\tan(t)=t

    and

    ln(1+\sin(t))=ln(1+t)=t

    we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of

    \frac{1}{\cos(x)^2}

    Which is 1.

    I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.
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    (Original post by james22)
    An alternative solution to the one given:

    Using well known aproximations and ignoring higher order terms (the limits go to 0 so we can assume higher roder terms don't matter) we have

    \tan(\sin(t))=\tan(t)=t

    and

    ln(1+\sin(t))=ln(1+t)=t

    we can then evaluate the integrals explicitly to get the original limit to be equal to the limit of

    \frac{1}{\cos(x)^2}

    Which is 0.

    I wonder if there is a better way of doing ths though? Both L'H and series expansions feel like cheating. A bit like solving an inequality by just using calculus.
    1/1 is 0?
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    (Original post by Occams Chainsaw)
    Yeah, I'm not really sure how use the latex yet. I'll look through the guide. I definitely didn't know how to display sigma. It's quite a nice tool, isn't it?!
    What I was trying to prove was what you interpreted. I now realise (after a rejuvinating 10 hours sleep and a Sunday roast) that I should have said I was trying to prove that the sum odd numbers is a square number!

    A nice little pattern I noticed is this:
    if I have an n*n box - or a perfect square
    xxxx
    xxxx
    xxxx
    xxxx

    in this case where n = 4, x = 16

    where x = a = b = c = d and we split it into (the only way I can describe it is 'mini'-squares' so the sum of the x-coordinate value for a will always equal the sum for a's y coordinate etc...)

    abcd
    abcc
    abbb
    aaaa

    So we have produced 7a + 5b + 3b + 1a = 16x
    or, 16 (a square number) = 7 + 5 + 3 + 1...

    Is that very obvious? It seems that way now I have written it down but I when I figured it out I felt pretty pleased with myself lol
    (n+1)^2=n^2+2n+1; In essence, to step from the nth square number to the n+1th, you must add the nth odd number. You can see this must be so as you say, the nth square number is a square, n*n. To get to the (n+1)th, you need to add a column of n, a row of n, and then one more (ie the 3 a's going downwards, the 3 a's at the bottom, then the one in the bottom left). Fairly clear, from a certain viewpoint. Otherwise, it's not entirely obvious that the sum of odd numbers should be square.
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    This may have been on here before so if it has please tell me and I'll delete it.

    Problem 346 *
    \displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx
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    Some number theory!

    Problem 347*

    Prove that  gcd(2^a - 1, 2^b - 1) = 2^{gcd(a,b)}- 1 .

    Problem 348**

    Prove that  \displaystyle \frac{x^2+1}{y^2-5} is never an integer for  x, y \in \mathbb{Z} .
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    (Original post by MW24595)
    Problem 348**

    Prove that  \displaystyle \frac{x^2+1}{y^2-5} is never an integer for  x, y \in \mathbb{Z} .
    This isn't true:

     \dfrac{2^2+1}{0^2-5}=-1
     \dfrac{x^2+1}{2^2-5}=-x^2-1

    If we change integer to positive integer:
    Spoiler:
    Show

    x^2+1 is not divisible by 4 (consider mod 4) and is is also not divisible by any primes of the form 4k-1 (factorise as (x+\mathrm{i})(x-\mathrm{i}), then since these don't have integer norms* they can be divisible by (Gaussian) primes with integer norms (ie. those of the form 4k-1)). However y^2-5 is either divisible by 4, is of the form 4k-1 (in which case it must have a factor of the form 4k-1) or is negative (which we have not allowed to make the statement true). Now that we have y^2-5 has a factor that x^2+1 doesn't the given result follows.

    *x=0 is a simple special case.

    The statement could also be made true by changing the denominator to y^2+3.
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    Solution 347

    WLOG b\geq a, let \delta be s.t. \delta |2^a-1 and \delta |2^b-1.

    Define a_{n+1}=\min (a_n, b_n-a_n) and b_{n+1}=\max (a_n,b_n-a_n) with a_1=a and b_1=b. Note that a_n\to 0 and b_n\to \gcd (a,b). For any n:

    \begin{aligned} &\delta |2^{b_n}-1\wedge \delta| 2^{a_n}-1\Rightarrow \delta | 2^{b_{n}}-2^{a_n}\Rightarrow \delta | 2^{b_n-a_n}-1\\& \Rightarrow \delta |2^{b_{n+1}}-1\wedge \delta| 2^{a_{n+1}}-1\Rightarrow \delta | 2^{b_{n+1}}-2^{a_{n+1}}\Rightarrow \cdots \\& \;\;\cdots \\&\quad \Rightarrow \delta |2^{\gcd (a,b)}-1\end{aligned}

    Obviously 2^{\gcd{(a,b)}}-1 divides both initial numbers and hence \delta_{\text{max}}= 2^{\gcd{(a,b)}}-1
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    Problem 349 / **

    Let h : \mathbb{R}^{\geq 0} \to \mathbb{R}^{+} be such that h'(x) > 2x for all x. Prove that

    \displaystyle \sum_{k=1}^{n} h(1 + F_k^2) > 4F_nF_{n+1}

    where F_n is the n-th Fibonacci.


    Problem 350 / **

    The Tribonacci numbers, T_n, are defined as

    T_0 = 0, T_1 = T_2 = 1 and T_{n+3} = T_{n+2} + T_{n+1} + T_{n} for n \geq 0.

    Find \displaystyle \sum_{k=1}^{n} T_k^2.
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    Solution 349

    I will use the identity

    \displaystyle\sum_{k=1}^n F_k^2=F_nF_{n+1}

    There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.

    Using standard results from analysis we have that h is increasing and that

    h(x)>x^2

    This is the best inequality we can have, as h can be made as close to x^2 as we like.

    Using both of these we have

    \displaystyle\sum_{i=1}^n h(1+(F_k)^2) >  \displaystyle\sum_{i=1}^n (1+(F_k)^2)^2 = \displaystyle\sum_{i=1}^n (1+2(F_k)^2+(F_k)^4)



\geq * \displaystyle\sum_{i=1}^n 4(F_k)^2 = 4F_nF_{n+1}

    You may not be happy with the starred inequality, but this is true because x^2+1>2x is immediate from (x-1)^2>0
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    (Original post by Pterodactyl)
    This may have been on here before so if it has please tell me and I'll delete it.

    Problem 346 *
    \displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx
    Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings.
    Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u
    Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
    After a bit of algebra bashing we obtain;
    \displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx = \int^b_a \frac{ax+bx-2ab}{2(a+b)(\sqrt{ax+bx-ab-x^2})}dx
    We now work on the RHS.
    We set  y = \frac{2x-a-b}{b-a}
    \displaystyle \int^1_{-1} \frac{(b^2-a^2)(y+1)}{4(a+b)(\sqrt{1-y^2})} dy
    Taking out the constants from the integral, we see our integral reduces down to
    \displaystyle \frac{b^2-a^2}{4(a+b)} \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy
    As y/(1-y^2) is obviously 0 over the interval as it is an odd function.
    This is now a well known integral;
    \displaystyle \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy = \arcsin(1) - \arcsin(-1) = \pi
    And thus
     I = \pi \frac{(b^2-a^2)}{4(a+b)}

    Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).

    I'll take a look at the tribonacci when I have some more time at a computer.
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    (Original post by jack.hadamard)
    Problem 350 / **

    The Tribonacci numbers, T_n, are defined as

    T_0 = 0, T_1 = T_2 = 1 and T_{n+3} = T_{n+2} + T_{n+1} + T_{n} for n \geq 0.

    Find \displaystyle \sum_{k=1}^{n} T_k^2.
    This is harder than I first thought. I have now failed with 3 methods, all failing because the reccurance relation has an odd number of terms (I think they would work with all ones with an even number of terms), the sums just don't all cancel as 1 is always left out.

    I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention.
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    (Original post by Llewellyn)
    Surprisingly tricky (I was barking down the wrong tree at 2am with inverse trig identities and far too many substitutions) but this has a very nice geometric link. (apologies if this is a repeat question, I haven't followed this thread much). Apologies also for the concise workings.
    Use IBP and set x - \frac{ab}{a+b} = v and arccos(...) = u
    Our uv term (arcos1-arccos1) can be safely set as 0 if I am to make any geometrical sense of this.
    After a bit of algebra bashing we obtain;
    \displaystyle \int^b_a \arccos \left(\frac{x}{\sqrt{\left(a+b \right)x - ab\right)}}\right)\ dx = \int^b_a \frac{ax+bx-2ab}{2(a+b)(\sqrt{ax+bx-ab-x^2})}dx
    We now work on the RHS.
    We set  y = \frac{2x-a-b}{b-a}
    \displaystyle \int^1_{-1} \frac{(b^2-a^2)(y+1)}{4(a+b)(\sqrt{1-y^2})} dy
    Taking out the constants from the integral, we see our integral reduces down to
    \displaystyle \frac{b^2-a^2}{4(a+b)} \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy
    As y/(1-y^2) is obviously 0 over the interval as it is an odd function.
    This is now a well known integral;
    \displaystyle \int^1_{-1} \frac{1}{\sqrt{1-y^2}} dy = \arcsin(1) - \arcsin(-1) = \pi
    And thus
     I = \pi \frac{(b^2-a^2)}{4(a+b)}

    Now (for the nice part) a sketch shows this describes a semicircle which intersects the x axis at a and b, and whereby the integrand describes the angle (for each x).

    I'll take a look at the tribonacci when I have some more time at a computer.
    I used a different second substitution so missed out on the nice geometric link.

    Problem 351 **
    \displaystyle \int^\infty_0 \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot4  }-\frac{x^7}{2\cdot4\cdot6}+... \right)\left(1+\frac{x^2}{2^2}+ \frac{x^4}{2^2\cdot4^2}+\frac{x^  6}{2^2\cdot4^2\cdot6^2}+... \right)\ dx

    Problem 352 */**
    \displaystyle \sum^\infty_{n=0} \dfrac{\left(-1\right)^n}{3n+1}
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    (Original post by Pterodactyl)
    Problem 351 **
    \displaystyle \int^\infty_0 \left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot4  }-\frac{x^7}{2\cdot4\cdot6}+... \right)\left(1+\frac{x^2}{2}+ \frac{x^4}{2^2\cdot4^2}+\frac{x^  6}{2^2\cdot4^2\cdot6^2}+... \right)\ dx
    Pardon me if I'm wrong, but should the second term of your second bracket be \frac{1}{2^2} x^2?

    Problem 353*/**
    For a positive integer n, let S_n be the total sum of the intervals of x such that \sin 4nx \geq \sin x in 0\leq x\leq \frac{\pi}{2}.

    Find \displaystyle \lim_{n\to\infty} S_n.

    Problem 354*/**

    Evaluate \displaystyle\lim_{n\to\infty} \left( \int_0^{\pi}\frac{\sin^2 nx}{\sin x} \ dx - \sum_{k=1}^{n}\frac{1}{k}\right).
 
 
 
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