Question

Find the constant term in the expansion of the binomial below:(x^3 + 2/x)^8A: 19843B: 17873C: 1792D: 1452

Answer 1

The binomial can be expanded using the formula:

`(a+b)^n=\sum ^n_(k\mathop=0)(n!)/((n-k)!k!)a^(n-k)b^k`

For the binomial (x³+2/x)⁸, we have:

n = 8

a = x³

b = 2/x

So, the constant term is the one for which:

`\mleft(x^3\mright)^(8-k)\mleft((2)/(x)\mright)^k=\text{ constant}`

This happens for k = 6:

`a^(n-k)b^k=(x^3)^(8-6)\mleft((2)/(x)\mright)^6=x^6\cdot(2^6)/(x^6)=2^6=64`

The, for k = 6, we have:

`(n!)/((n-k)!k!)=(8!)/(2!6!)=(8\cdot7\cdot6!)/(2\cdot1\cdot6!)=(56)/(2)=28`

Thus, the constant term in the expansion of the binomial is:

`(n!)/((n-k)!k!)a^(n-k)b^k=28\cdot64=1792`

Therefore, the answer is 1792.