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secondary 4 | A Maths
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hi can someone pls help me with this qn,thankyou!
Date Posted: 2 years ago
Views: 183
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Hope this helps
Date Posted:
2 years ago
Actually no.
The values of r in nCr goes from 0 to n where r are integers.
The first term in the expansion involves r = 0.
The second term involves r = 1.
The third term involves r = 2.
The kth term involves r = k - 1.
These are called the general "r + 1"th term.
So for the sixth term, r = 5.
The values of r in nCr goes from 0 to n where r are integers.
The first term in the expansion involves r = 0.
The second term involves r = 1.
The third term involves r = 2.
The kth term involves r = k - 1.
These are called the general "r + 1"th term.
So for the sixth term, r = 5.
I realised that. Sorry for the mistake. The correct answer should be n=15 and 3003/32. Thanks for pointing out.
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Date Posted:
2 years ago
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