Key Concepts Used

This problem involves finding the coefficient of a specific power of $x$ in a sum of binomial expansions. To solve it efficiently, we will utilize the following fundamental concepts from combinatorics and algebra:

1. Binomial Theorem: The expansion of a binomial $(a+b)^n$ is given by the formula: $(a+b)^n = \sum_{r=0}^n {}^{n}C_r a^{n-r} b^r$ For $(1+x)^n$, the general term is ${}^{n}C_r x^r$, and thus, the coefficient of $x^r$ in $(1+x)^n$ is ${}^{n}C_r$.

2. Symmetry Property of Combinations: The number of ways to choose $r$ items from a set of $n$ items is the same as choosing $n-r$ items from the set. Mathematically: ${}^{n}C_r = {}^{n}C_{n-r}$ This property is often useful for simplifying expressions involving binomial coefficients.

3. Hockey-stick Identity (Summation Identity for Binomial Coefficients): This powerful identity allows us to sum a series of binomial coefficients: $\sum_{i=r}^n {}^{i}C_r = {}^{r}C_r + {}^{r+1}C_r + \dots + {}^{n}C_r = {}^{n+1}C_{r+1}$ It's named for its appearance on Pascal's Triangle.

4. Geometric Progression (GP) Sum (Alternative Approach): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of the first $N$ terms of a GP is $S_N = a \frac{1 - r^N}{1 - r}$, where $a$ is the first term and $r$ is the common ratio.

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Step-by-step Solution

We need to find the coefficient of ${x^{301}}$ in the series: $S = (1 + x)^{500} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$

Step 1: Decompose the Series and Identify Terms Contributing to $x^{301}$

The given series is a sum of several terms. We need to find the coefficient of $x^{301}$ in each individual term and then sum them up. Let's look at a general term in the series: The $k^{th}$ term (starting with $k=0$ for the first term) can be written as $T_k = x^k (1+x)^{500-k}$.

To find the coefficient of $x^{301}$ in $T_k$, we recognize that $x^k$ is already present. Therefore, we need to find the coefficient of $x^{301-k}$ from the binomial expansion of $(1+x)^{500-k}$. According to the Binomial Theorem, the coefficient of $x^{301-k}$ in $(1+x)^{500-k}$ is ${}^{500-k}C_{301-k}$.

For a term to contribute to $x^{301}$, the power of $x$ being sought in the binomial expansion, $(301-k)$, must be non-negative. This means $301-k \ge 0 \Rightarrow k \le 301$. Additionally, the power of $x$ being sought must not exceed the power of the binomial, i.e., $301-k \le 500-k$, which simplifies to $301 \le 500$, always true. Also, the exponent of the binomial $500-k$ must be non-negative, so $k \le 500$. Combining these conditions, $k$ can range from $0$ up to $301$.

Let's list the coefficients for the first few terms and the last contributing term:

• For $k=0$ (term $x^0(1+x)^{500}$): We need the coefficient of $x^{301}$ in $(1+x)^{500}$, which is ${}^{500}C_{301}$.
• For $k=1$ (term $x^1(1+x)^{499}$): We need the coefficient of $x^{300}$ in $(1+x)^{499}$, which is ${}^{499}C_{300}$.
• For $k=2$ (term $x^2(1+x)^{498}$): We need the coefficient of $x^{299}$ in $(1+x)^{498}$, which is ${}^{498}C_{299}$.
• ...
• For $k=301$ (term $x^{301}(1+x)^{199}$): We need the coefficient of $x^0$ in $(1+x)^{199}$, which is ${}^{199}C_0$.

Summing these up, the total coefficient of $x^{301}$ is: ${}^{500}{C_{301}} + {}^{499}{C_{300}} + {}^{498}{C_{299}}\, + \,...\, + \,{}^{199}{C_0}$

Step 2: Apply the Symmetry Property of Combinations

The sum obtained in Step 1 is not directly in the form of the Hockey-stick identity because the lower index of the combinations is not constant. To prepare it for the identity, we use the symmetry property ${}^{n}C_r = {}^{n}C_{n-r}$. This transformation will make the lower index constant.

Let's transform each term in the sum:

• ${}^{500}C_{301} = {}^{500}C_{500-301} = {}^{500}C_{199}$
• ${}^{499}C_{300} = {}^{499}C_{499-300} = {}^{499}C_{199}$
• ${}^{498}C_{299} = {}^{498}C_{498-299} = {}^{498}C_{199}$
• ...
• ${}^{199}C_0 = {}^{199}C_{199-0} = {}^{199}C_{199}$

Now, the sum becomes: ${}^{500}{C_{199}} + {}^{499}{C_{199}} + {}^{498}{C_{199}}\, + \,...\, + \,{}^{199}{C_{199}}$ To match the standard form of the Hockey-stick identity, it's often clearer to write this sum in ascending order of the upper index: ${}^{199}{C_{199}} + {}^{200}{C_{199}} + \dots + {}^{499}{C_{199}} + {}^{500}{C_{199}}$

Step 3: Apply the Hockey-stick Identity

Now the sum is in the perfect form for the Hockey-stick Identity, which states: $\sum_{i=r}^n {}^{i}C_r = {}^{r}C_r + {}^{r+1}C_r + \dots + {}^{n}C_r = {}^{n+1}C_{r+1}$

Comparing our sum $\sum_{i=199}^{500} {}^{i}C_{199}$ with the identity:

• The constant lower index is $r = 199$.
• The upper index $i$ ranges from $199$ to $500$, so the maximum upper index is $n=500$.

Applying the Hockey-stick Identity, we replace $n$ with $500$ and $r$ with $199$: ${}^{199}{C_{199}} + {}^{200}{C_{199}} + \dots + {}^{500}{C_{199}} = {}^{500+1}C_{199+1}$ $= {}^{501}C_{200}$

Alternative Method: Summing the Geometric Progression

The given series can also be recognized as a Geometric Progression (GP). $S = (1 + x)^{500} + x{(1 + x)^{499}} + {x^2}{(1 + x)^{498}}\, + \,...\, + \,{x^{500}}$

Let's identify the elements of this GP:

• First term ($a$) = $(1+x)^{500}$
• Common ratio ($r$) = $\frac{x(1+x)^{499}}{(1+x)^{500}} = \frac{x}{1+x}$
• Number of terms ($N$) = $501$ (from $k=0$ to $k=500$ for $x^k$)

The sum of a GP is given by $S_N = a \frac{1 - r^N}{1 - r}$. Substituting the values into the GP sum formula: $S = (1+x)^{500} \frac{1 - \left(\frac{x}{1+x}\right)^{501}}{1 - \frac{x}{1+x}}$ Simplify the numerator: $1 - \frac{x^{501}}{(1+x)^{501}} = \frac{(1+x)^{501} - x^{501}}{(1+x)^{501}}$ Simplify the denominator: $1 - \frac{x}{1+x} = \frac{1+x-x}{1+x} = \frac{1}{1+x}$

Now substitute these back into the sum expression: $S = (1+x)^{500} \frac{\frac{(1+x)^{501} - x^{501}}{(1+x)^{501}}}{\frac{1}{1+x}}$ $S = (1+x)^{500} \cdot \frac{(1+x)^{501} - x^{501}}{(1+x)^{501}} \cdot (1+x)$ $S = \frac{(1+x)^{500} \cdot (1+x) \cdot ((1+x)^{501} - x^{501})}{(1+x)^{501}}$ $S = (1+x)^{501} - x^{501}$

Now, we need to find the coefficient of $x^{301}$ in $S = (1+x)^{501} - x^{501}$.

• From the term $(1+x)^{501}$, the coefficient of $x^{301}$ is ${}^{501}C_{301}$ (by Binomial Theorem).
• The term $-x^{501}$ does not contain $x^{301}$ (since $501 \neq 301$), so its contribution to the coefficient of $x^{301}$ is $0$.

Thus, the coefficient of $x^{301}$ in the entire series is ${}^{501}C_{301}$. Using the symmetry property of combinations, we can write this as: ${}^{501}C_{301} = {}^{501}C_{501-301} = {}^{501}C_{200}$ Both methods yield the same result, confirming our answer.

Step 4: Final Answer

The coefficient of ${x^{301}}$ in the given series is ${}^{501}C_{200}$.

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Important Tips and Common Mistakes

• Recognizing Series Patterns: Always be on the lookout for common series patterns like Arithmetic Progression (AP), Geometric Progression (GP), or specific binomial series. Simplifying the sum first (as shown in the alternative method) can often lead to a much cleaner solution.
• Careful with Indices: When applying binomial identities or extracting coefficients, pay close attention to the indices ($n$ and $r$). A small error in indexing can lead to a completely different result.
• Hockey-stick Identity Requirements: Remember that for the Hockey-stick Identity, the lower index ($r$) must be constant across all terms in the sum, and the upper index ($i$) must be consecutive, starting from $r$. If the sum doesn't perfectly match, use properties like ${}^nC_r = {}^nC_{n-r}$ to transform the terms.
• Understanding "Coefficient of $x^k$": When a term is $x^j(1+x)^m$, and you need the coefficient of $x^k$, you are looking for the coefficient of $x^{k-j}$ in $(1+x)^m$. This is ${}^mC_{k-j}$. Ensure that $k-j$ is non-negative and less than or equal to $m$.

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Summary and Key Takeaway

This problem demonstrates the power of combining multiple combinatorial and algebraic techniques. The most direct approach involves breaking down the series into individual terms, applying the Binomial Theorem to find the relevant coefficient from each, and then using the symmetry property of combinations to transform the sum into a form suitable for the Hockey-stick Identity. An alternative, more elegant approach is to first recognize the series as a Geometric Progression, sum it up, and then extract the required coefficient. Both methods consistently lead to the correct answer, ${}^{501}C_{200}$. Mastering these identities and problem-solving strategies is crucial for efficiently tackling complex problems in combinatorics.