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secondary 3 | A Maths
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Date Posted: 3 years ago
Views: 182
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A is 2/x
B is x^2/k
Term rth= nCr . A^(n-r) . B^r
n is 6
(2/X)^(6-r) . (X^2/k)^r = 2^(6-r) /k^r . X^(2r -6+r)
Constant term means index is zero
2r -6 +r is zero then r is 2
Then you sub r=2 to 6C2 (use pascal triangle it is 6) . 2^4/k^2 = 60( I forgot got to read the question)
Then solv k is +/-
B is x^2/k
Term rth= nCr . A^(n-r) . B^r
n is 6
(2/X)^(6-r) . (X^2/k)^r = 2^(6-r) /k^r . X^(2r -6+r)
Constant term means index is zero
2r -6 +r is zero then r is 2
Then you sub r=2 to 6C2 (use pascal triangle it is 6) . 2^4/k^2 = 60( I forgot got to read the question)
Then solv k is +/-
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