Hi,
How would you do parts (b) and (c) for this question?
Please could you also show the steps to how you got the answer as well?
Thank you!![]()
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Shanahey
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- 11-05-2015 12:44
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Kevin De Bruyne
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- 11-05-2015 12:48
(Original post by Shanahey)
Hi,
How would you do parts (b) and (c) for this question?
Please could you also show the steps to how you got the answer as well?
Thank you!
It helps to think about what went on from u2 to u3 to u4, and what u5 will be. -
Shanahey
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- 11-05-2015 12:50
(Original post by SeanFM)
(b) and (c) rely on (a). What did you get for those values?
It helps to think about what went on from u2 to u3 to u4, and what u5 will be.
I got U2 as 2/3, U3 as -4 and u4 as 3 -
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- 11-05-2015 12:53
Without further calculation, can you see what u5 would be? (You can calculate it if you want but hopefully you spot something else). -
Shanahey
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- 11-05-2015 12:56
(Original post by SeanFM)
Brilliant
Without further calculation, can you see what u5 would be? (You can calculate it if you want but hopefully you spot something else). -
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- 11-05-2015 12:59
(Original post by Shanahey)
U5 would be 2/3? -
Shanahey
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- 11-05-2015 13:03
(Original post by SeanFM)
Bingo. Can you see the pattern? -
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- 11-05-2015 13:08
So U1 = a, U2 = b, U3 = c (for the first 3 numbers), U4 = a .... (for the numbers 4-6)... what will U61 be? -
Shanahey
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- 11-05-2015 13:16
(Original post by SeanFM)
Well, the terms seem to appear in sets (which you'll have worked out as sets of 3 - a,b,c).
So U1 = a, U2 = b, U3 = c (for the first 3 numbers), U4 = a .... (for the numbers 4-6)... what will U61 be? -
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- 11-05-2015 13:54
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