Key Concepts and Formulas

This problem primarily utilizes the following 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 $n$ terms of a G.P. with first term $a$ and common ratio $r$ is given by: $S_n = a \frac{r^n - 1}{r - 1} \quad \text{or} \quad S_n = a \frac{1 - r^n}{1 - r}$
2. Binomial Theorem: For any non-negative integer $n$, the binomial expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ Specifically, for $(1+x)^n$, the expansion is: $(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$ The coefficient of $x^k$ in $(1+x)^n$ is ${ }^nC_k$ or $\binom{n}{k}$.
3. Binomial Coefficient Identity: ${ }^nC_r = { }^nC_{n-r}$. This property allows for alternative representations of binomial coefficients.
4. Hockey-stick Identity (or Christmas Stocking Identity): For $r \le n$, the sum of binomial coefficients of the form $\sum_{i=r}^n { }^iC_r = { }^{n+1}C_{r+1}$.

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Step-by-Step Solution

1. Identify the Series as a Geometric Progression (G.P.)

The given expression is a sum of terms: $S = x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\ldots+x^{54}(1+x)^{46}$

• First term ($a$): Observe the first term of the series, which is $x^2(1+x)^{98}$.
• Common ratio ($r$): To find the common ratio, divide the second term by the first term: $r = \frac{x^3(1+x)^{97}}{x^2(1+x)^{98}} = \frac{x}{1+x}$ We can verify this with the next pair of terms: $\frac{x^4(1+x)^{96}}{x^3(1+x)^{97}} = \frac{x}{1+x}$. This confirms it's a G.P.
• Number of terms ($n$): The power of $x$ starts from 2 and goes up to 54. The number of terms is $54 - 2 + 1 = 53$.

2. Calculate the Sum of the G.P.

Using the formula $S_n = a \frac{1 - r^n}{1 - r}$: $S = x^2(1+x)^{98} \cdot \frac{1 - \left(\frac{x}{1+x}\right)^{53}}{1 - \frac{x}{1+x}}$

Now, simplify the terms:

• Denominator: $1 - \frac{x}{1+x} = \frac{(1+x) - x}{1+x} = \frac{1}{1+x}$
• Numerator: $1 - \left(\frac{x}{1+x}\right)^{53} = 1 - \frac{x^{53}}{(1+x)^{53}} = \frac{(1+x)^{53} - x^{53}}{(1+x)^{53}}$

Substitute these back into the sum formula: $S = x^2(1+x)^{98} \cdot \frac{\frac{(1+x)^{53} - x^{53}}{(1+x)^{53}}}{\frac{1}{1+x}}$ To simplify, multiply by the reciprocal of the denominator: $S = x^2(1+x)^{98} \cdot \frac{(1+x)^{53} - x^{53}}{(1+x)^{53}} \cdot (1+x)$ Combine the powers of $(1+x)$: $S = x^2 \frac{(1+x)^{98} (1+x)}{(1+x)^{53}} \left((1+x)^{53} - x^{53}\right)$ $S = x^2 (1+x)^{99-53} \left((1+x)^{53} - x^{53}\right)$ $S = x^2 (1+x)^{46} \left((1+x)^{53} - x^{53}\right)$ Distribute $x^2(1+x)^{46}$: $S = x^2 (1+x)^{46} (1+x)^{53} - x^2 (1+x)^{46} x^{53}$ $S = x^2 (1+x)^{99} - x^{55} (1+x)^{46}$

3. Find the Coefficient of $x^{70}$ in the Sum

We need to find the coefficient of $x^{70}$ in each part of the simplified sum:

• Part 1: Coefficient of $x^{70}$ in $x^2 (1+x)^{99}$ To get $x^{70}$ from $x^2 (1+x)^{99}$, we need the coefficient of $x^{70-2} = x^{68}$ from the binomial expansion of $(1+x)^{99}$. Using the binomial theorem, the coefficient of $x^{68}$ in $(1+x)^{99}$ is ${ }^{99}C_{68}$.

• Part 2: Coefficient of $x^{70}$ in $-x^{55} (1+x)^{46}$ To get $x^{70}$ from $-x^{55} (1+x)^{46}$, we need the coefficient of $x^{70-55} = x^{15}$ from the binomial expansion of $-(1+x)^{46}$. Using the binomial theorem, the coefficient of $x^{15}$ in $(1+x)^{46}$ is ${ }^{46}C_{15}$. Since the term is negative, the coefficient is $-{ }^{46}C_{15}$.

Combining these, the coefficient of $x^{70}$ in the entire sum is: ${ }^{99}C_{68} - { }^{46}C_{15}$

4. Compare with the Given Form and Find $p+q$

The problem states that the coefficient is ${ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$. By direct comparison, we can identify $p=68$ and $q=15$. Therefore, $p+q = 68 + 15 = 83$.

Alternative Representation using ${ }^nC_r = { }^nC_{n-r}$

We can also use the identity ${ }^nC_r = { }^nC_{n-r}$ to find other possible values for $p$ and $q$.

• ${ }^{99}C_{68} = { }^{99}C_{99-68} = { }^{99}C_{31}$
• ${ }^{46}C_{15} = { }^{46}C_{46-15} = { }^{46}C_{31}$

So, the coefficient of $x^{70}$ can also be expressed as: ${ }^{99}C_{31} - { }^{46}C_{31}$ In this case, $p=31$ and $q=31$. Therefore, $p+q = 31 + 31 = 62$.

The problem asks for "a possible value of $p+q$". Our calculations yield two possible values: 83 and 62.

• Option (A) 61
• Option (B) 83
• Option (C) 55
• Option (D) 68

Based on our derivation, $p+q=83$ is a possible value, which corresponds to option (B). The provided "Correct Answer: A" (61) does not align with a standard mathematical derivation of the given series sum. If 61 were the correct answer, it would imply a different initial sum or a misinterpretation of the binomial coefficients in the problem statement. However, adhering strictly to the properties of G.P. and binomial theorem, 83 (and 62) are the derived results.

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Summary and Key Takeaway

The problem involves first identifying and summing a geometric progression, then extracting the coefficient of a specific power of $x$ using the binomial theorem. It's crucial to correctly identify the first term, common ratio, and number of terms in the G.P. and then carefully apply the binomial expansion. Remember that binomial coefficients can often be expressed in two forms (${ }^nC_r$ and ${ }^nC_{n-r}$), leading to multiple "possible values" for sums like $p+q$. In this case, standard derivation leads to $p+q=83$ or $p+q=62$.

Tip: Always double-check the limits of the series and the indices when applying summation formulas or binomial identities to avoid off-by-one errors. Also, be aware that problems sometimes present options that don't perfectly align with all possible interpretations, requiring careful re-evaluation of the problem statement.