Understanding the Binomial Expansion
The question asks us to identify the largest coefficient in the expansion of the binomial expression \((1 + x)^{100}\).
We can use the Binomial Theorem to expand expressions of the form \((a+b)^n\). The theorem states:
\((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)In our specific case, \(a = 1\), \(b = x\), and \(n = 100\). Substituting these values into the Binomial Theorem formula, we get:
\((1+x)^{100} = \sum_{k=0}^{100} \binom{100}{k} 1^{100-k} x^k\)Since \(1\) raised to any power is \(1\), the formula simplifies to:
\((1+x)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^k\)This expansion looks like:
\(\binom{100}{0}x^0 + \binom{100}{1}x^1 + \binom{100}{2}x^2 + \dots + \binom{100}{k}x^k + \dots + \binom{100}{100}x^{100}\)The coefficients in this expansion are the binomial coefficients, denoted by \(\binom{100}{k}\), where \(k\) ranges from \(0\) to \(100\).
Finding the Greatest Coefficient
To find the greatest coefficient, we need to understand how binomial coefficients \(\binom{n}{k}\) behave. The general property of binomial coefficients for a fixed \(n\) is that they increase as \(k\) increases from \(0\) up to \(n/2\), and then decrease as \(k\) increases from \(n/2\) to \(n\). The maximum value occurs when \(k\) is closest to \(n/2\).
In our case, \(n = 100\). Therefore, \(n/2 = 100/2 = 50\).
The coefficient \(\binom{100}{k}\) will be greatest when \(k = 50\). This means the greatest coefficient is \(\binom{100}{50}\).
Let's examine the coefficients corresponding to the options provided:
- The coefficient of \(x^{100}\) is \(\binom{100}{100}\). We know that \(\binom{n}{n} = 1\). So, this coefficient is \(1\).
- The coefficient of \(x^{99}\) is \(\binom{100}{99}\). Using the property \(\binom{n}{k} = \binom{n}{n-k}\), we have \(\binom{100}{99} = \binom{100}{100-99} = \binom{100}{1}\). We know that \(\binom{n}{1} = n\). So, this coefficient is \(100\).
- The coefficient of \(x^{51}\) is \(\binom{100}{51}\).
- The coefficient of \(x^{50}\) is \(\binom{100}{50}\).
Comparing the coefficients:
We know that \(\binom{100}{50}\) is the central binomial coefficient and is the largest. The coefficients are symmetric around \(k=50\). Thus, \(\binom{100}{50} > \binom{100}{51}\) (since \(50\) is exactly \(n/2\) and \(51\) is slightly further away). Also, \(\binom{100}{50}\) is significantly larger than \(\binom{100}{99}\) and \(\binom{100}{100}\).
Therefore, the greatest coefficient in the expansion of \((1 + x)^{100}\) is the coefficient of \(x^{50}\).
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