The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

STEP 2017 Solutions

Scroll to see replies

Original post by 1 8 13 20 42
..For the last part, one route that works is:

k(n+k−1)(n+k−2)=n−1n+k−1−n−2n+k−2 \displaystyle \frac{k}{(n + k - 1)(n + k - 2)} = \dfrac{n-1}{n+k-1}-\dfrac{n-2}{n+k-2}

=1n+k−1+(n−2)(1n+k−1−1n+k−2)=\displaystyle \dfrac{1} {n+k-1} + (n-2)\left( \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2}\right)

But ∑1n+k−1\displaystyle \sum \dfrac{1}{n+k-1} diverges as it's just ∑1/n\sum 1/n with a different starting index, and summing 1n+k−1−1n+k−2\displaystyle \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2} telescopes (and therefore converges). So the overall sum must diverge.

This feels a bit "tough" for STEP II (not that it's necessarily difficult but it's a very first year analysis approach), but I'm not seeing anything better.

Edit: Just to say that I would rather show the sum diverges by considering k/((n+k-1)(n+k-2)) and observing that for k > n, the denominator is less than 4k^2 and so k/((n+k-1)(n+k-2)) is > 1/(4k) and therefore diverges by comparison with 1/k. But that doesn't use partial fractions as demanded by the question.
(edited 6 years ago)
Original post by DFranklin
For the last part, one route that works is:

k(n+k−1)(n+k−2)=n−1n+k−1−n−2n+k−2 \displaystyle \frac{k}{(n + k - 1)(n + k - 2)} = \dfrac{n-1}{n+k-1}-\dfrac{n-2}{n+k-2}

=1n+k−1+(n−2)(1n+k−1−1n+k−2)=\displaystyle \dfrac{1} {n+k-1} + (n-2)\left( \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2}\right)

But ∑1n+k−1\displaystyle \sum \dfrac{1}{n+k-1} diverges as it's just ∑1/n\sum 1/n with a different starting index, and summing 1n+k−1−1n+k−2\displaystyle \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2} telescopes (and therefore converges). So the overall sum must diverge.

This feels a bit "tough" for STEP II (not that it's necessarily difficult but it's a very first year analysis approach), but I'm not seeing anything better.


I didn't know a level students were supposed to know 1/n diverged.


Posted from TSR Mobile
Original post by physicsmaths
I didn't know a level students were supposed to know 1/n diverged.It's specifically given as information in the question.
Original post by DFranklin
For the last part, one route that works is:

k(n+k−1)(n+k−2)=n−1n+k−1−n−2n+k−2 \displaystyle \frac{k}{(n + k - 1)(n + k - 2)} = \dfrac{n-1}{n+k-1}-\dfrac{n-2}{n+k-2}

=1n+k−1+(n−2)(1n+k−1−1n+k−2)=\displaystyle \dfrac{1} {n+k-1} + (n-2)\left( \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2}\right)

But ∑1n+k−1\displaystyle \sum \dfrac{1}{n+k-1} diverges as it's just ∑1/n\sum 1/n with a different starting index, and summing 1n+k−1−1n+k−2\displaystyle \dfrac{1}{n+k-1}-\dfrac{1}{n+k-2} telescopes (and therefore converges). So the overall sum must diverge.

This feels a bit "tough" for STEP II (not that it's necessarily difficult but it's a very first year analysis approach), but I'm not seeing anything better.

Edit: Just to say that I would rather show the sum diverges by considering k/((n+k-1)(n+k-2)) and observing that for k > n, the denominator is less than 4k^2 and so k/((n+k-1)(n+k-2)) is > 1/(4k) and therefore diverges by comparison with 1/k. But that doesn't use partial fractions as demanded by the question.


Ah yeah that's not too bad, guess it makes more sense to do the partial fractions for the whole expression as opposed to just splitting the fraction with numerator 1, that seemed to go nowhere for me. But certainly it is a bit creatively demanding for STEP, especially when most people haven't had that much experience playing with series and limits. I definitely prefer that kind of approach as well. Mind if I just copy it in and credit you?
Original post by DFranklin
It's specifically given as information in the question.


Oh I see
Q12:

(i)



(ii)



(iii)




(iv)

(edited 6 years ago)
Original post by 1 8 13 20 42
Ah yeah that's not too bad, guess it makes more sense to do the partial fractions for the whole expression as opposed to just splitting the fraction with numerator 1, that seemed to go nowhere for me. But certainly it is a bit creatively demanding for STEP, especially when most people haven't had that much experience playing with series and limits. I definitely prefer that kind of approach as well. Mind if I just copy it in and credit you?
Fine by me to just copy it in.

It feels like both STEP I and STEP II have had questions with a bit more "university analysis" than usual.
Original post by DFranklin
Q12:

(i)



(ii)



(iii)




(iv)



Very nice solutions. I thought it was a fairly straightforward question.
Any estimation about the boundaries? How much for an S...
Hi, could someone help me approximate my marks, please? :smile:

1) All parts apart from induction. Induction was attempted but I believe I got the wrong value for A.
2) Correct part a and got an expression for n+4 in terms of n but did not simplify.
3) Only drew y = x for first graph and did not draw the upper and lower parts for sin(y)=1/2 sin(x) graph. Also did not draw cos(y) = 1/2 sin(x) correctly. Apart from that it's all correct.
6) Waffled through proof by induction but it kind of made sense. Did not do any more of question.
7) Parts i, ii, iii all correct. Not able to do iv and graph for g(x) is incorrect.

I would really appreciate it if someone can help me to approximate a mark, thank you!
Reply 210
Avatar for abc653
abc653
Original post by Farhan.Hanif93
STEP II Q2

Solution



Actually, I think the condition ∣b−a∣<2|b-a|<2 is extraneous. Indeed, 4=2(a2+b2)=(b+a)2+(b−a)2≥(b−a)24=2(a^2+b^2)=(b+a)^2+(b-a)^2\geq (b-a)^2. Equality is achieved if and only if a=1,b=−1a=1, b=-1 or a=−1,b=1a=-1, b=1. However, both cases along with xn2+(b−a)xn+1=0x_{n}^2+(b-a)x_{n}+1=0 give us that xn+b=0x_{n}+b=0, which is impossible according to the problem statement.
Reply 211
Avatar for SM-
SM-
6
How many marks would part iii of 5 and finding C in 6 be?
Original post by SM-
How many marks would part iii of 5 and finding C in 6 be?


Don't know about Q5 but finding C in Q6 seems like it would be at least 8 marks.
Hey guys, can anyone estimate my marks and an approximate grade?

1. Fully completed
2. Messed up a bit here, got a+b = 0 as a necessary condition but i did not write x^2 + (b-a)x -1 is not 0, and i got the conditions for period 4 however, again i did not say x^2 + (b-a)x -1 is not 0
4. Fully completed
5. Fully completed
6. Did the first 2 parts however could not do the last part
7. Did part 2 and part 3 out of the four parts (i know, an odd selection to do)

Thank you to anyone who estimates a mark and a grade :biggrin:
Reply 214
Avatar for Zacken
Zacken
OP
22
Original post by Manniman
Hey guys, can anyone estimate my marks and an approximate grade?

1. Fully completed
2. Messed up a bit here, got a+b = 0 as a necessary condition but i did not write x^2 + (b-a)x -1 is not 0, and i got the conditions for period 4 however, again i did not say x^2 + (b-a)x -1 is not 0
4. Fully completed
5. Fully completed
6. Did the first 2 parts however could not do the last part
7. Did part 2 and part 3 out of the four parts (i know, an odd selection to do)

Thank you to anyone who estimates a mark and a grade :biggrin:


19 + 16 + 19 + 19 + 12 + 6 -- 91, should be a high 1/low S.
I've put up a solution to 9 here:

https://www.thestudentroom.co.uk/showpost.php?p=72114070&postcount=200

but it could do with checking. It also needs a diagram, but I don't have time at the moment.
Q11

(i)



(ii)

(edited 6 years ago)
Original post by solC
Since there isn't a solution for it, and I have nothing better to do, I'm gonna post my *attempt* at Q11 and maybe someone could use it construct a complete solution?I've posted my attempt at this just above.
Looking at your argument, I think a serioous error occurs when you try to calculate the maximum height using conservation of energy - you seem to assume all the KE is converted to PE at this point, but the particle will still have horizontal velocity, so non-zero KE. I haven't really tried to follow what you wrote beyond this, but I'm not sure much can be salvaged.
Excuse me, gentlemen. I know nearly nothing about STEP and its evaluation criteria, so I would really appreciate if someone could estimate my marks and grade for STEP II. Thanks.

Q1) Fully completed;
Q2) Completed except (ii);
Q3) Did just (i);
Q4) Fully completed;
Q6) Everything except determining C (misunderstood the question);
Q7) Proved first inequality in (i), did (ii) and found first limit in (iii) :frown:
Original post by Zacken
19 + 16 + 19 + 19 + 12 + 6 -- 91, should be a high 1/low S.


Thanks bro, I just needed a first so i guess that's good. On to STEP III!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you