Key Concepts and Formulas

This problem primarily utilizes two fundamental mathematical concepts:

1. Geometric Progression (G.P.): A sequence of numbers 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 G.P. with first term $a$ and common ratio $r$ is given by: $S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{if } r \neq 1$ (This form is generally preferred when $|r| < 1$, but it is algebraically equivalent to $\frac{a(r^n - 1)}{r - 1}$ and can be used for any $r \neq 1$.)

2. Binomial Theorem: For any non-negative integer $n$, the binomial expansion of $(A+B)^n$ is given by: $(A+B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k$ The general term, or the $(k+1)^{\text{th}}$ term, in the expansion of $(A+B)^n$ is $T_{k+1} = \binom{n}{k} A^{n-k} B^k$. This term gives the coefficient of $B^k$ multiplied by $A^{n-k}$.

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Step-by-Step Solution

1. Identify the Geometric Progression (G.P.) The given expression is a sum of terms: $(5 + x)^{500} + x(5 + x)^{499} + x^2(5 + x)^{498} + \,\,.....\,\, + \,\,x^{500}$ To determine if this is a G.P., we examine the ratio of consecutive terms.

• Let the first term be $T_1 = (5+x)^{500}$.
• The second term is $T_2 = x(5+x)^{499}$.
• The third term is $T_3 = x^2(5+x)^{498}$.

Calculate the ratio of $T_2$ to $T_1$: $\frac{T_2}{T_1} = \frac{x(5+x)^{499}}{(5+x)^{500}} = \frac{x}{5+x}$ Now, calculate the ratio of $T_3$ to $T_2$: $\frac{T_3}{T_2} = \frac{x^2(5+x)^{498}}{x(5+x)^{499}} = \frac{x}{5+x}$ Since the ratio of consecutive terms is constant, the given expression is indeed a Geometric Progression.

2. Determine the Parameters of the G.P. From our observation in step 1, we can identify the following parameters:

• First term ($a$): The first term in the series is $a = (5+x)^{500}$.
• Common ratio ($r$): The constant ratio we found is $r = \frac{x}{5+x}$.
• Number of terms ($n$): Let's list the terms in a way that helps count them. The terms are: $(5+x)^{500} \cdot x^0$ $(5+x)^{499} \cdot x^1$ $(5+x)^{498} \cdot x^2$ $\vdots$ $(5+x)^0 \cdot x^{500} \quad \text{(since } x^{500} \text{ can be written as } x^{500}(5+x)^0)$ The power of $(5+x)$ decreases from $500$ down to $0$. The number of terms is given by (highest power - lowest power + 1), so $500 - 0 + 1 = 501$ terms. Thus, $n = 501$.

3. Calculate the Sum of the G.P. We use the formula for the sum of $n$ terms of a G.P.: $S_n = \frac{a(1 - r^n)}{1 - r}$. Substitute the values of $a$, $r$, and $n$: $S_{501} = \frac{(5+x)^{500} \left(1 - \left(\frac{x}{5+x}\right)^{501}\right)}{1 - \frac{x}{5+x}}$

Now, let's simplify this expression step-by-step:

• Simplify the denominator: $1 - \frac{x}{5+x} = \frac{(5+x) - x}{5+x} = \frac{5}{5+x}$

• Simplify the term inside the numerator's bracket: $1 - \left(\frac{x}{5+x}\right)^{501} = 1 - \frac{x^{501}}{(5+x)^{501}} = \frac{(5+x)^{501} - x^{501}}{(5+x)^{501}}$

• Substitute the simplified terms back into the sum formula: $S_{501} = \frac{(5+x)^{500} \left(\frac{(5+x)^{501} - x^{501}}{(5+x)^{501}}\right)}{\frac{5}{5+x}}$

• Perform the division by multiplying by the reciprocal of the denominator: $S_{501} = (5+x)^{500} \left(\frac{(5+x)^{501} - x^{501}}{(5+x)^{501}}\right) \times \frac{5+x}{5}$

• Combine the terms and simplify exponents: $S_{501} = \frac{(5+x)^{500} \cdot ( (5+x)^{501} - x^{501} ) \cdot (5+x)}{5 \cdot (5+x)^{501}}$ Notice that $(5+x)^{500} \cdot (5+x) = (5+x)^{501}$. So, the expression becomes: $S_{501} = \frac{(5+x)^{501} \cdot ( (5+x)^{501} - x^{501} )}{5 \cdot (5+x)^{501}}$ We can cancel out $(5+x)^{501}$ from the numerator and denominator: $S_{501} = \frac{(5+x)^{501} - x^{501}}{5}$ This can be rewritten as: $S_{501} = \frac{1}{5} \left( (5+x)^{501} - x^{501} \right)$

4. Find the Coefficient of $x^{101}$ in the Sum We need to find the coefficient of $x^{101}$ in $S_{501} = \frac{1}{5} \left( (5+x)^{501} - x^{501} \right)$.

• Analyze the term $(5+x)^{501}$: We need the coefficient of $x^{101}$ from the expansion of $(5+x)^{501}$. Using the Binomial Theorem, for $(A+B)^n$, the term with $B^k$ is $\binom{n}{k} A^{n-k} B^k$. Here, $A=5$, $B=x$, $n=501$, and we are looking for $x^{101}$, so $k=101$. The term will be: $\binom{501}{101} (5)^{501-101} (x)^{101} = \binom{501}{101} 5^{400} x^{101}$. So, the coefficient of $x^{101}$ in $(5+x)^{501}$ is $\binom{501}{101} 5^{400}$.

• Analyze the term $-x^{501}$: This term is simply $-x^{501}$. It contains $x$ raised to the power of $501$. Since we are looking for the coefficient of $x^{101}$ (and $101 \neq 501$), this term does not contribute to the coefficient of $x^{101}$. Its coefficient for $x^{101}$ is $0$.

• Combine the results for the total sum: The coefficient of $x^{101}$ in $S_{501}$ is $\frac{1}{5}$ times the coefficient of $x^{101}$ from the combined terms $(5+x)^{501} - x^{501}$. $\text{Coefficient of } x^{101} = \frac{1}{5} \left( \left( \text{Coefficient of } x^{101} \text{ in } (5+x)^{501} \right) - \left( \text{Coefficient of } x^{101} \text{ in } x^{501} \right) \right)$ $= \frac{1}{5} \left( \binom{501}{101} 5^{400} - 0 \right)$ $= \frac{1}{5} \binom{501}{101} 5^{400}$ $= \binom{501}{101} 5^{400-1}$ $= \binom{501}{101} 5^{399}$

Final Answer: The coefficient of $x^{101}$ in the given expression is $\binom{501}{101} 5^{399}$.

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Tips and Common Mistakes to Avoid

• Verify G.P. properties: Always double-check that the series is indeed a G.P. by calculating the common ratio.
• Accurate Term Counting: A frequent error is miscounting the number of terms in a series. Remember that if powers run from $P_{\text{max}}$ to $P_{\text{min}}$, the number of terms is $P_{\text{max}} - P_{\text{min}} + 1$.
• Algebraic Simplification: Be meticulous with algebraic manipulations, especially when dealing with fractions and exponents in the sum calculation. Even a small error in simplification can lead to a completely different result.
• Binomial Theorem General Term: Ensure you correctly identify $A$, $B$, $n$, and $k$ when applying the Binomial Theorem. Remember that for the term containing $B^k$, you use $\binom{n}{k}$.
• Irrelevant Terms: When looking for a specific power of $x$, any terms in the overall expression that do not contain that power (or contain a higher/lower power exclusively) can be safely ignored for that part of the calculation, as their coefficient for the desired power will be zero.

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Summary and Key Takeaway

This problem is a classic example that combines the properties of Geometric Progressions with the Binomial Theorem. The main challenge is to first recognize the given series as a G.P. and then use its sum formula to simplify the entire expression. Once simplified to the form $\frac{1}{5} \left( (5+x)^{501} - x^{501} \right)$, the task transforms into finding the coefficient of $x^{101}$ in a binomial expansion, which is a direct application of the Binomial Theorem. This problem highlights the importance of recognizing underlying series patterns and applying the correct expansion formulas.