The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Sequence and series. Show and find unknown a

Hi, can someone explain the approach I should be taking for this question.

Sequences Isn’t one of my strong points. I didn’t recall any of it from GCSE level before starting this chapter, so I had to revisit to get the basics.

As far as I can tell I basically have to take the formula given.

In terms of n / n+1. To find the a. Can someone confirm this or am I completely wrong?

Part b
CA1E1420-43F7-4E86-8DFA-B84AAE98683D.jpeg

I have already solved part a. That’s fairly simple
(edited 2 years ago)
So un = n2n
Then what would un+1 be?
Reply 2
Avatar for KingRich
KingRich
OP
14
Original post by flaurie
So un = n2n
Then what would un+1 be?

Ahh. I was being simple. I hadn’t thought of b. Being an extension of part a.

I believe I understand it now.

C6A6B7FB-5F69-403E-AED5-287438F273E4.jpeg

Apologies if the thought process isn’t very clear
(edited 2 years ago)
B21022C4-A5CA-40EA-BB13-9934AFF1B771.jpeg145.5KB
Reply 3
Avatar for mqb2766
mqb2766
21
Thats not quite right, you have to show the ratio is a + a/n, rather than assuming it and subbing in the values for n=1 and n=2.

Youve got the right starting point for u_(n+1)/u_n, so how can you simplify that expression to put it in the form they want? The values for n=1 and n=2 are just to get you warmed up. They are not relevant for this question part.
(edited 2 years ago)
Reply 4
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
Thats not quite right, you have to show the ratio is a + a/n, rather than assuming it and subbing in the values for n=1 and n=2.

Youve got the right starting point for u_(n+1)/u_n, so how can you simplify that expression to put it in the form they want? The values for n=1 and n=2 are just to get you warmed up. They are not relevant for this question part.

Oh, :frown: lol

Erm, do you mean simplifying nth term function?
FFAE0167-4CED-4DA2-9533-9E29282764F4.jpeg
Reply 5
Avatar for mqb2766
mqb2766
21
Yes. From answer spotting, the 2^n type terms must disappear/cancel.
Reply 6
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
Yes. From answer spotting, the 2^n type terms must disappear/cancel.

Okay, from the top of my head from similar questions in the past. To cancel out 2^n. I want to multiply by 2^n/2^n. However, showing the calculations to confirm this, is a little trouble some.
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by KingRich
Okay, from the top of my head from similar questions in the past. To cancel out 2^n. I want to multiply by 2^n/2^n. However, showing the calculations to confirm this, is a little trouble some.

Assuming youve split 2^(n+1) into 2*2^n, then you have 2^n on both the numerator and the denominator so they simply cancel.
Reply 8
Avatar for KingRich
KingRich
OP
14
I

Original post by mqb2766
Assuming youve split 2^(n+1) into 2*2^n, then you have 2^n on both the numerator and the denominator so they simply cancel.

Ah, believe I’ve figured it out lol.
1D4F3355-7297-40BE-B0E4-6BAD24B33D1B.jpeg
Reply 9
Avatar for mqb2766
mqb2766
21
Original post by KingRich
I


Ah, believe I’ve figured it out lol.
1D4F3355-7297-40BE-B0E4-6BAD24B33D1B.jpeg

Yes, thats right. You obviously get a=2 again, but this time youve shown the result for an arbitrary n.

Note there is no need to multiply by 2^n/2^n at the start. The existing 2^n terms simply cancel in the fraction.
(edited 2 years ago)
Reply 10
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
Yes, thats right. You obviously get a=2 again, but this time youve shown the result for an arbitrary n.

Note there is no need to multiply by 2^n/2^n at the start. The existing 2^n terms simply cancel in the fraction.

Thank you for this. You always come to my rescue and help me understand things better.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you