Question

The fourth term in the expansion of `(x^2 + b)^11` is `55/72x^16`. What is the value of `b`?

Answer 1

`b = 1/6`

Explanation:

When expanding `(a + b)^n`, we typically go in descending powers of `a` and ascending powers of `b`.

There will be `11+ 1` terms in this expansion, since we must count the 0th term.

The first will be `"something"x^22`, the next `"something else"x^20`. If we continue like this, we see that the 4th term will be of the form `"something"x^16`.

Now, the formula for the nth term in a binomial expansion is `t_(k + 1) = color(white)(two)_nC_k a^(n - k)b^k`. So if we're looking for `t_4`, then

`k + 1 = 4 -> k = 3`

We know `t_4`, so we will be solving for b.

`55/72x^16 = color(white)(two)_11C_3 (x^2)^(11 - 3) b^3`

We compute `color(white)(two)_11C_3` using `color(white)(two)_nC_r = (n!)/((n - r)!r!)`.

`color(white)(two)_11C_3 = (11!)/((11 - 3)!3!) = 165`

Putting all of this back together:

`55/72x^16 = 165x^16b^3`

Our goal is to solve for `b`.

`(55/72x^16)/(165x^16) = b^3`

`55/11880= b^3`

You can use a calculator to simplify to

`b = 1/6`

Hopefully this helps!