Binomial Theorem
Coefficient in Binomial Expansion
Grade 11
Question:
<p>If \((1+x)^{2016}\left[\dfrac{\left(\dfrac{x}{1+x}\right)^{2017}-1}{\dfrac{1}{1+x}-1}\right] = \sum_{i=0}^{2016} a_i x^i\), then \(a_{17}\) equals:</p>
<p>(1) \(\dfrac{2017!}{17! \cdot 2000!}\)</p>
<p>(2) \(\dfrac{2016!}{17! \cdot 1999!}\)</p>
<p>(3) \(\dfrac{2017!}{16! \cdot 2001!}\)</p>
<p>(4) \(\dfrac{2017!}{17! \cdot 2000!}\)</p>
Step-by-Step Solution
Key Concept: Simplify the complex fraction inside the brackets by recognizing that the denominator equals -x/(1+x), then use binomial expansion to identify the coefficient of x^17 in the resulting polynomial expression.
<p><strong>Step 1:</strong> Simplify the denominator in brackets: $\frac{1}{1+x} - 1 = \frac{1-(1+x)}{1+x} = \frac{-x}{1+x}$</p><p><strong>Step 2:</strong> Simplify the entire bracket expression: $\frac{\left(\frac{x}{1+x}\right)^{2017}-1}{\frac{-x}{1+x}} = \frac{\left(\frac{x}{1+x}\right)^{2017}-1}{\frac{-x}{1+x}}$</p><p><strong>Step 3:</strong> Let $y = \frac{x}{1+x}$. Then: $\frac{y^{2017}-1}{\frac{-x}{1+x}} = \frac{(1+x)(y^{2017}-1)}{-x}$</p><p><strong>Step 4:</strong> Since $y = \frac{x}{1+x}$, we have $y^{2017} = \frac{x^{2017}}{(1+x)^{2017}}$. The expression becomes: $(1+x)^{2016} \cdot \frac{(1+x)}{-x}\left(\frac{x^{2017}}{(1+x)^{2017}}-1\right)$</p><p><strong>Step 5:</strong> Simplifying: $(1+x)^{2016} \cdot \frac{1}{-x}\left(\frac{x^{2017}-(1+x)^{2017}}{(1+x)^{2017}}\right) = \frac{1}{-x}\left(x^{2017}-(1+x)^{2017}\right)$</p><p><strong>Step 6:</strong> Expanding $(1+x)^{2017} = \sum_{k=0}^{2017}\binom{2017}{k}x^k$, the coefficient of $x^{18}$ in $x^{2017}-(1+x)^{2017}$ divided by $-x$ gives the coefficient of $x^{17}$.</p><p><strong>Step 7:</strong> $a_{17} = -\binom{2017}{18}$ (the coefficient of $x^{18}$ in $(1+x)^{2017}$ with negative sign)</p><p>∴ Answer: A</p>
Correct Answer: A