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:

$(a+b)^n=\sum ^n_{k\mathop=0}\frac{n!}{(n-k)!k!}a^{n-k}b^k$

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:

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

This happens for k = 6:

$a^{n-k}b^k=(x^3)^{8-6}\mleft(\frac{2}{x}\mright)^6=x^6\cdot\frac{2^6}{x^6}=2^6=64$

The, for k = 6, we have:

$\frac{n!}{(n-k)!k!}=\frac{8!}{2!6!}=\frac{8\cdot7\cdot6!}{2\cdot1\cdot6!}=\frac{56}{2}=28$

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

$\frac{n!}{(n-k)!k!}a^{n-k}b^k=28\cdot64=1792$

Therefore, the answer is 1792.