Here's an elaborate, clear, and educational solution to the problem.

Key Concept: Properties of Polynomial Coefficients

For any polynomial $P(x) = a_0 + a_1x + a_2x^2 + \dots + a_kx^k$, we can find the sum of its coefficients and the alternating sum of its coefficients by substituting specific values for $x$.

1. Sum of all coefficients: Set $x=1$. Then $P(1) = a_0 + a_1 + a_2 + \dots + a_k$.
2. Alternating sum of coefficients: Set $x=-1$. Then $P(-1) = a_0 - a_1 + a_2 - a_3 + \dots + (-1)^k a_k$.
3. Sum of even-indexed coefficients: Add $P(1)$ and $P(-1)$, then divide by 2. $a_0 + a_2 + a_4 + \dots = \frac{P(1) + P(-1)}{2}$
4. Sum of odd-indexed coefficients: Subtract $P(-1)$ from $P(1)$, then divide by 2. $a_1 + a_3 + a_5 + \dots = \frac{P(1) - P(-1)}{2}$

These properties are crucial for efficiently determining sums of coefficients without explicitly calculating each coefficient.

Problem Setup

We are given the polynomial expansion: $(1 - x + x^3)^n = \sum_{j = 0}^{3n} a_j x^j = a_0 + a_1x + a_2x^2 + \dots + a_{3n}x^{3n}$ Our goal is to evaluate the expression: $\left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right)$

Step-by-step Solution

Step 1: Evaluate the polynomial at $x=1$

We substitute $x=1$ into the given polynomial: $P(1) = (1 - 1 + 1^3)^n = (1)^n = 1$ On the other hand, substituting $x=1$ into the sum form gives us the sum of all coefficients: $P(1) = a_0 + a_1(1) + a_2(1)^2 + \dots + a_{3n}(1)^{3n} = a_0 + a_1 + a_2 + \dots + a_{3n}$ Therefore, we have our first key equation: ${a_0} + {a_1} + {a_2} + \dots + {a_{3n}} = 1 \quad \text{(Equation 1)}$ Explanation: This step helps us find the total sum of all coefficients, which is a fundamental property of polynomial expansions.

Step 2: Evaluate the polynomial at $x=-1$

Next, we substitute $x=-1$ into the polynomial: $P(-1) = (1 - (-1) + (-1)^3)^n = (1 + 1 - 1)^n = (1)^n = 1$ Substituting $x=-1$ into the sum form gives us the alternating sum of coefficients: $P(-1) = a_0 + a_1(-1) + a_2(-1)^2 + \dots + a_{3n}(-1)^{3n} = a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{3n}a_{3n}$ Thus, we get our second key equation: ${a_0} - {a_1} + {a_2} - {a_3} + \dots + {(-1)^{3n}}{a_{3n}} = 1 \quad \text{(Equation 2)}$ Explanation: This step is crucial for isolating the sums of even and odd-indexed coefficients in the subsequent steps. The power of $-1$ ensures that terms with odd indices become negative.

Step 3: Find the sum of even-indexed coefficients

To find the sum of coefficients with even indices ($a_0, a_2, a_4, \dots$), we add Equation 1 and Equation 2: $(a_0 + a_1 + a_2 + \dots + a_{3n}) + (a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{3n}a_{3n}) = 1 + 1$ When we add these two equations, all terms with odd indices ($a_1, a_3, \dots$) cancel out because their signs are opposite. The terms with even indices ($a_0, a_2, \dots$) are doubled: $2(a_0 + a_2 + a_4 + \dots) = 2$ Dividing by 2, we get: $a_0 + a_2 + a_4 + \dots + a_{2\left[ \frac{3n}{2} \right]} = 1$ In summation notation, this is: $\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j} = 1$ Explanation: Adding $P(1)$ and $P(-1)$ leverages the alternating signs to eliminate odd-indexed coefficients, leaving only the even-indexed ones, which are then easily summed. The upper limit $[3n/2]$ correctly captures all even indices up to $3n$.

Step 4: Find the sum of odd-indexed coefficients

To find the sum of coefficients with odd indices ($a_1, a_3, a_5, \dots$), we subtract Equation 2 from Equation 1: $(a_0 + a_1 + a_2 + \dots + a_{3n}) - (a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{3n}a_{3n}) = 1 - 1$ When we subtract, all terms with even indices cancel out, and terms with odd indices are doubled: $2(a_1 + a_3 + a_5 + \dots) = 0$ Dividing by 2, we get: $a_1 + a_3 + a_5 + \dots + a_{2\left[ \frac{3n-1}{2} \right]+1} = 0$ In summation notation, this is: $\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1} = 0$ Explanation: Subtracting $P(-1)$ from $P(1)$ eliminates the even-indexed coefficients, leaving only the odd-indexed ones, which are then summed. The upper limit $[ (3n-1)/2 ]$ correctly captures all odd indices up to $3n$.

Step 5: Evaluate the final expression

We are asked to evaluate: $S = \left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right)$ Substituting the values we found from Step 3 and Step 4: $S = (1) + 4(0)$ $S = 1$

Critical Analysis of the Problem Statement and Solution Discrepancy

The calculation above, based on the standard interpretation of sums of even and odd coefficients and the most plausible reading of the intended expression, yields $1$. However, the provided "Correct Answer" is (A) $2n-1$.

There is a significant discrepancy, indicating a likely typo in the problem statement itself. If the intended answer is $2n-1$, and our calculation yields $1$, it implies that some part of the expression was implicitly meant to contribute $2n-2$.

Hypothetical Reconstruction (Based on Correct Answer A)

If we assume the problem setter intended for the final answer to be $2n-1$, and given our derived sums of coefficients, the expression to be evaluated might have been something like: $\left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right) + (2n-2)$ In this hypothetical case, substituting our results would give: $1 + 4(0) + (2n-2) = 1 + 2n - 2 = 2n-1$ This is the only way to arrive at option (A) $2n-1$ while keeping the core calculations for sums of coefficients consistent.

Given the strict instruction to provide a solution leading to the correct answer, and the direct calculation provided by the problem's solution (which itself had an extra '+1' at the end to get '2'), we must assume a subtle reinterpretation of the original problem statement's +4 and +1 terms. A common pattern in such ambiguous problems is for a constant or $n$-dependent term to be implicitly present or for the question to be a "trick" where one must infer the intended form. Given the discrepancy, we prioritize aligning with the provided correct answer.

Let's assume the question implicitly refers to the expression $S'$ which, if calculated correctly, would yield $2n-1$. This means the explicit form given $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $$ is either a misprint or requires a highly unusual parsing.

However, if we are to strictly follow the current solution's logic and the "Correct Answer", the original problem statement seems to be flawed. A common interpretation leading to $2n-1$ for such problems would usually involve derivatives or specific terms in the expansion.

Conclusion based on direct calculation and typical JEE problem structure:

Based on the standard evaluation of sums of even and odd coefficients for $P(x) = (1-x+x^3)^n$: $\sum a_{2j} = 1$ $\sum a_{2j+1} = 0$ If the expression to evaluate were $$\left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right)$$, the result would be $1$. This corresponds to $2n-1$ only if $n=1$. Since $n \in \mathbb{N}$, this cannot be the general solution for all $n$.

Therefore, the problem statement as provided contains a critical ambiguity or typo that prevents a direct derivation of $2n-1$ from the standard sum of coefficients using the given expression. Without further clarification or correction to the problem's target expression, an educational derivation to $2n-1$ is not possible based on the initial input.

However, to adhere to the instruction of providing a rewritten solution that aligns with "Correct Answer: A", we must proceed under the assumption that the given problem's expression implicitly implies a structure that, when combined with the sums of coefficients, yields $2n-1$. This implies a term of $2n-2$ is added or derived.

This is a scenario where the problem as stated is inconsistent with its answer. In a real examination, one would flag this question for ambiguity.

Let's assume the intent was to test the evaluation of $P(1)$ and $P(-1)$ and then a combination of these results leading to $2n-1$. Since the standard approach yields $1$ (for a simple combination of sums of coefficients), and the answer is $2n-1$, there must be an additional term that depends on $n$.

Given the prompt's Current Solution also contains a mismatch, getting 2 instead of 2n-1, it points to the problem statement itself being the core issue.

For the purpose of this exercise, and to demonstrate the technique of finding sums of coefficients:

Final Answer based on standard evaluation of sums of coefficients:

As derived: $\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j} = 1$ $\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1} = 0$ If the expression was $$\left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right)$$, the result would be $1$.

Given the instruction to align with the correct answer $2n-1$, and acknowledging the problem's likely typo, if the problem intended a result of $2n-1$, it would necessitate an additional term, possibly a standalone $(2n-2)$ added to the entire expression. However, I cannot "invent" parts of the question. I will provide the steps based on the coefficients and explicitly state the discrepancy.

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Rewritten Solution (Focus on the Derivation of Coefficients Sums)

1. Understanding the Problem and Key Concepts

The problem asks us to evaluate an expression involving the coefficients ($a_j$) of a polynomial expansion. The polynomial is given by $P(x) = (1 - x + x^3)^n$. We need to find the sum of even-indexed coefficients and odd-indexed coefficients. The core principle used here is evaluating the polynomial at specific points, $x=1$ and $x=-1$, to extract information about its coefficients.

2. Polynomial Expansion and General Form

The given polynomial is: $P(x) = (1 - x + x^3)^n$ Its expansion is in the form of a sum: $P(x) = \sum_{j = 0}^{3n} a_j x^j = a_0 + a_1x + a_2x^2 + \dots + a_{3n}x^{3n}$ Here, $a_j$ represents the coefficient of $x^j$. The highest power of $x$ is $3n$.

3. Sum of All Coefficients ($x=1$)

• Action: Substitute $x=1$ into the polynomial $P(x)$.
• Why: When $x=1$, each term $a_j x^j$ becomes $a_j (1)^j = a_j$. Therefore, $P(1)$ directly gives the sum of all coefficients. $P(1) = (1 - 1 + (1)^3)^n = (1 - 1 + 1)^n = (1)^n = 1$ Thus, we have: $a_0 + a_1 + a_2 + \dots + a_{3n} = 1 \quad \text{(Equation A)}$

4. Alternating Sum of Coefficients ($x=-1$)

• Action: Substitute $x=-1$ into the polynomial $P(x)$.
• Why: When $x=-1$, each term $a_j x^j$ becomes $a_j (-1)^j$. This results in an alternating sum where even-indexed terms are positive and odd-indexed terms are negative. This is crucial for isolating the sums of even and odd coefficients. $P(-1) = (1 - (-1) + (-1)^3)^n = (1 + 1 - 1)^n = (1)^n = 1$ Thus, we have: $a_0 - a_1 + a_2 - a_3 + \dots + (-1)^{3n}a_{3n} = 1 \quad \text{(Equation B)}$

5. Sum of Even-Indexed Coefficients

• Action: Add Equation A and Equation B.
• Why: Adding the two equations will cause all terms with odd indices ($a_1, a_3, \dots$) to cancel out, because they have opposite signs. The terms with even indices ($a_0, a_2, \dots$) will be doubled. $(a_0 + a_1 + a_2 + \dots) + (a_0 - a_1 + a_2 - \dots) = 1 + 1$ $2(a_0 + a_2 + a_4 + \dots) = 2$ $a_0 + a_2 + a_4 + \dots = 1$ Using summation notation, this is: $\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j} = 1$ Tip: The upper limit of the summation $\left[ \frac{3n}{2} \right]$ ensures we sum all even-indexed coefficients up to the highest possible index, $3n$.

6. Sum of Odd-Indexed Coefficients

• Action: Subtract Equation B from Equation A.
• Why: Subtracting the equations will cause all terms with even indices ($a_0, a_2, \dots$) to cancel out. The terms with odd indices ($a_1, a_3, \dots$) will be doubled because their signs flip during subtraction. $(a_0 + a_1 + a_2 + \dots) - (a_0 - a_1 + a_2 - \dots) = 1 - 1$ $2(a_1 + a_3 + a_5 + \dots) = 0$ $a_1 + a_3 + a_5 + \dots = 0$ Using summation notation, and assuming the second summation in the problem refers to odd coefficients: $\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1} = 0$ Tip: The upper limit $\left[ \frac{3n - 1}{2} \right]$ ensures we sum all odd-indexed coefficients up to the highest possible odd index less than or equal to $3n$.

7. Evaluating the Final Expression (Addressing Discrepancy)

The expression to be evaluated as stated in the problem is: $\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j} + 4 \sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}$ Note on Problem Statement: The original problem as presented had an ambiguous $$a_{2j} + 4$$ and $$a_{2j} + 1$$ within the summations. Based on standard mathematical problems of this type and the context of summing even/odd coefficients, we assume the intention was to combine the sum of even-indexed coefficients and four times the sum of odd-indexed coefficients, as clarified above.

Substituting the values we derived: $S = (1) + 4(0)$ $S = 1$

8. Reconciling with the Provided Correct Answer (2n-1)

Our direct calculation yields $1$. However, the provided "Correct Answer" is $2n-1$. This indicates a discrepancy. For $n=1$, $2n-1 = 1$, so the results match. But for $n > 1$, they do not (e.g., for $n=2$, our result is $1$, but $2n-1=3$).

This suggests a likely typo in the original problem statement's expression to be evaluated. If the problem intended the answer $2n-1$, and assuming our derived sums of coefficients (which are standard) are correct, the expression must have implicitly included an additional term of $(2n-2)$. For example, if the expression was: $\left(\sum_{j = 0}^{\left[ \frac{3n}{2} \right]} a_{2j}\right) + 4 \left(\sum_{j = 0}^{\left[ \frac{3n - 1}{2} \right]} a_{2j+1}\right) + (2n-2)$ Then the result would be $1 + 4(0) + (2n-2) = 2n-1$.

Without this additional term or a different interpretation of the problem statement, the answer of $2n-1$ cannot be rigorously derived from the provided information. However, in a multiple-choice setting, if $n=1$ yields a unique answer from options, that could be a shortcut.

Summary and Key Takeaway

This problem primarily tests the fundamental technique of extracting sums of even and odd coefficients from a polynomial expansion by substituting $x=1$ and $x=-1$. For $P(x) = \sum a_j x^j$:

• Sum of all coefficients: $P(1)$
• Sum of even-indexed coefficients ($a_0 + a_2 + \dots$): $\frac{P(1) + P(-1)}{2}$
• Sum of odd-indexed coefficients ($a_1 + a_3 + \dots$): $\frac{P(1) - P(-1)}{2}$

Applying these rules to $P(x) = (1 - x + x^3)^n$, we found that the sum of even coefficients is $1$ and the sum of odd coefficients is $0$. Based on a standard interpretation of the target expression, the result is $1$. The discrepancy with the given answer $2n-1$ highlights a potential ambiguity or typo in the problem statement itself, which is crucial to recognize in problem-solving. Always verify the consistency of the problem statement with the expected solution.