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maths question - sequences

ill post the questions asap
Reply 1
image-c149628a-157c-42d1-b5ce-8778d82b7be55216858482629738323-compressed.jpg.jpeg question 10
Reply 2
image-da995109-16d1-4472-9589-205472c6f7f15164243016895821419-compressed.jpg.jpeg challenge question
Original post by sqrt of 5
question 10


Nothing stopping you from applying the technique of Q8b to deduce the value of xx hence state what the first term is (2x2x)
Reply 4
Original post by RDKGames
Nothing stopping you from applying the technique of Q8b to deduce the value of xx hence state what the first term is (2x2x)

ok... got it
Original post by sqrt of 5
challenge question


The differences become the same at 3rd layer of comparisons, so the nth term will be a cubic of the form:

an=An3+Bn2+Cn+Da_n = An^3 + Bn^2 + Cn + D

You can use the facts that a1=3,a2=20,a3=63,a4=144a_1 = 3, a_2 = 20, a_3 = 63, a_4 = 144 to deduce four equations in A,B,C,DA,B,C,D and solve them simultaneously.
Reply 6
Original post by RDKGames
The differences become the same at 3rd layer of comparisons, so the nth term will be a cubic of the form:

an=An3+Bn2+Cn+Da_n = An^3 + Bn^2 + Cn + D

You can use the facts that a1=3,a2=20,a3=63,a4=144a_1 = 3, a_2 = 20, a_3 = 63, a_4 = 144 to deduce four equations in A,B,C,DA,B,C,D and solve them simultaneously.

i was close! I did an^3+bn^2+c
thank you :smile:

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