(i) Given that the constant term in the binomial expansion of
is 7, find the value of the positive constant k.
(ii) Using the value of k found in part (i), show that there is no constant term in the expansion of
(i) For the first expansion, we recall the general term of a binomial expansion, which is
and this simplifies to
Hence, the term independent of x (constant term) is when 8 – 4r = 0
or when r = 2
therefore, the constant term is 8C2 (-k)2 = 7
28k2 = 7
k = ½
(ii) This is a tricky one. At first glance, one might think that the question is wrong. Afterall, if the constant term of the binomial expansion is 7 and if it is multiplied by a 1 from the expression (1 + x4), that will surely give a constant of 7. So why would there not be a constant term? Well, that is because we also need to look at what the x4 term multiplies with. If the corresponding x-4 term is -7, then when the two constants are added together, it is zero!
So, let us now find out what the coefficient of x-4 is in the binomial expansion.
For the term in x-4, then 8 – 4r = -4
4r = 12
r = 3
Substitute r = 3 into the expression, we get the coefficient of x-4 as 8C3(-½)3 = 56(-1/8) = -7!
Hence, we have shown that there would not be a constant term in the expansion.
Maths is cool, isn't it??
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