Finding the Largest Coefficient in Binomial Expansion

The question asks for the largest coefficient in the expansion of \((x + 1)^{20}\).

We use the Binomial Theorem to expand \((a + b)^n\), which is given by:

\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are the binomial coefficients.

In our case, we have \((x + 1)^{20}\), so \(a = x\), \(b = 1\), and \(n = 20\).

The expansion is:

\((x + 1)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^{20-k} 1^k = \sum_{k=0}^{20} \binom{20}{k} x^{20-k}\)

The coefficients in this expansion are \(\binom{20}{k}\) for \(k = 0, 1, 2, \ldots, 20\).

The values of the binomial coefficients \(\binom{n}{k}\) for a fixed \(n\) increase as \(k\) goes from \(0\) up to \(\frac{n}{2}\) (or the integers closest to \(\frac{n}{2}\)) and then decrease. For an even value of \(n\), the largest coefficient occurs at the middle term, where \(k = \frac{n}{2}\).

Here, \(n = 20\), which is an even number. The largest coefficient will occur when \(k = \frac{20}{2} = 10\).

The largest coefficient is therefore \(\binom{20}{10}\).

Let's calculate \(\binom{20}{10}\) using the formula:

\(\binom{20}{10} = \frac{20!}{10!(20-10)!} = \frac{20!}{10!10!}\)

This can also be written as \(\frac{(20)!}{(10!)^2}\).

Comparing this result with the given options:

• Option 1: \(\frac{(20)!}{(10!)^2}\)
• Option 2: \(\frac{(20)!}{6!4!}\)
• Option 3: \(\frac{(20)!}{5!4!}\)
• Option 4: \(\frac{(20)!}{10!5!}\)

The largest coefficient is \(\frac{(20)!}{(10!)^2}\), which matches Option 1.