Key Concepts and Formulas

This problem primarily uses two fundamental concepts:

1. Geometric Progression (GP) Sum Formula: For a geometric series with first term $a$, common ratio $r$, and $n$ terms, the sum $S_n$ is given by: $S_n = a \frac{1 - r^n}{1 - r} \quad (\text{when } r \ne 1)$
2. Binomial Theorem: For any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. For the special case $(1+x)^n$, the general term (coefficient of $x^k$) is $\binom{n}{k}$.

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Step-by-Step Solution

1. Identify the Series as a Geometric Progression The given series is: $P = (1 + x)^{2016} + x(1 + x)^{2015} + x^2 (1 + x)^{2014} + \dots + x^{2016}$ Let's analyze the terms:

• First term ($T_1$): $(1+x)^{2016}$
• Second term ($T_2$): $x(1+x)^{2015}$
• Third term ($T_3$): $x^2(1+x)^{2014}$ ...
• Last term ($T_{2017}$): $x^{2016}$ (since $(1+x)^0=1$)

We observe that each successive term is obtained by multiplying the previous term by a constant factor. This indicates a Geometric Progression.

2. Determine the First Term, Common Ratio, and Number of Terms

• First term ($a$): As identified above, $a = (1+x)^{2016}$.

• Common Ratio ($r$): The common ratio is the factor by which each term is multiplied to get the next term. We can find it by dividing the second term by the first term: $r = \frac{T_2}{T_1} = \frac{x(1+x)^{2015}}{(1+x)^{2016}} = \frac{x}{1+x}$ Self-check: Let's also check with $T_3/T_2$: $\frac{x^2(1+x)^{2014}}{x(1+x)^{2015}} = \frac{x}{1+x}$ The common ratio is indeed $r = \frac{x}{1+x}$.

• Number of terms ($n$): The powers of $x$ in the terms range from $x^0$ (in the first term, implicitly $(1+x)^{2016} \cdot x^0$) to $x^{2016}$. The powers of $(1+x)$ range from $(1+x)^{2016}$ to $(1+x)^0$. There are $2016 - 0 + 1 = 2017$ terms in total. So, $n = 2017$.

3. Apply the GP Sum Formula Now, we use the formula for the sum of a GP, $S_n = a \frac{1 - r^n}{1 - r}$: $P = (1+x)^{2016} \cdot \frac{1 - \left(\frac{x}{1+x}\right)^{2017}}{1 - \frac{x}{1+x}}$

4. Simplify the Expression for P First, simplify the denominator: $1 - \frac{x}{1+x} = \frac{(1+x) - x}{1+x} = \frac{1}{1+x}$ Now substitute this back into the sum expression: $P = (1+x)^{2016} \cdot \frac{1 - \frac{x^{2017}}{(1+x)^{2017}}}{\frac{1}{1+x}}$ $P = (1+x)^{2016} \cdot (1+x) \cdot \left(1 - \frac{x^{2017}}{(1+x)^{2017}}\right)$ $P = (1+x)^{2017} \cdot \left(\frac{(1+x)^{2017} - x^{2017}}{(1+x)^{2017}}\right)$ $P = \frac{(1+x)^{2017} \cdot ((1+x)^{2017} - x^{2017})}{(1+x)^{2017}}$ $P = (1+x)^{2017} - x^{2017}$

5. Find the Coefficient of $x^{17}$ We are given that $P = \sum_{i=0}^{2016} a_i x^i$. This means we need to find the coefficient of $x^{17}$ in the expression for $P$, but only considering terms up to $x^{2016}$.

Our simplified expression for $P$ is $(1+x)^{2017} - x^{2017}$. Since we are looking for $a_{17}$ (the coefficient of $x^{17}$), we only need to consider the terms up to $x^{2016}$. The term $-x^{2017}$ does not contribute to $a_{17}$ because its power is $2017$, which is greater than $17$. Therefore, $a_{17}$ is the coefficient of $x^{17}$ in the expansion of $(1+x)^{2017}$.

Using the Binomial Theorem for $(1+x)^n$, the coefficient of $x^k$ is $\binom{n}{k}$. Here, $n=2017$ and $k=17$. So, the coefficient of $x^{17}$ is $\binom{2017}{17}$.

6. Calculate the Binomial Coefficient $\binom{2017}{17} = \frac{2017!}{17!(2017-17)!} = \frac{2017!}{17!2000!}$

This matches option (A).

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Tips and Common Mistakes

• Careful with GP identification: Ensure the ratio between consecutive terms is constant.
• Accurate count of terms: A common error is miscounting the number of terms, especially when powers start from zero or a different number. In this case, $x^0$ to $x^{2016}$ makes it $2017$ terms.
• Algebraic Simplification: Be meticulous with algebraic steps, particularly when dealing with fractions within fractions.
• Binomial Expansion Range: Remember that the problem asks for coefficients up to $x^{2016}$. If a term like $-x^{2017}$ appears, it only affects coefficients for powers equal to or greater than $2017$.

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Summary and Key Takeaway

This problem beautifully combines two important mathematical concepts: the sum of a Geometric Progression and the Binomial Theorem. The main strategy was to first simplify the given series into a more manageable form using the GP sum formula, and then apply the Binomial Theorem to extract the required coefficient. The key takeaway is to recognize underlying series patterns (like GP) and simplify them before performing further operations like finding specific coefficients.