Key Concepts and Formulas

This problem involves two fundamental concepts from the Binomial Theorem and its generalizations:

1. Number of terms in a multinomial expansion: For an expansion of the form $(x_1 + x_2 + \dots + x_k)^n$, the total number of terms is given by the formula: $\binom{n + k - 1}{k - 1} \quad \text{or equivalently} \quad \binom{n + k - 1}{n}$ In our specific case, we have a trinomial (meaning $k=3$) expansion of the form $(a+b+c)^n$. Therefore, the number of terms will be $\binom{n+3-1}{3-1} = \binom{n+2}{2}$.

2. Sum of the coefficients in an expansion: To find the sum of the coefficients of all terms in any polynomial expansion (e.g., $(ax+by+cz)^n$ or similar forms), we simply substitute the value 1 for all the variables present in the base expression. This works because when variables are 1, their powers remain 1, effectively summing up only the coefficients.

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Step-by-Step Solution

Step 1: Determine the value of 'n'

• Understanding the Expression: We are given the expression ${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$ This expression is a trinomial raised to the power $n$, where the terms are $1$, $-\frac{2}{x}$, and $\frac{4}{x^2}$. Here, $k=3$ (number of terms in the base).

• Applying the Number of Terms Formula: According to the formula for a trinomial expansion, the number of terms is $\binom{n+2}{2}$. We are given that the number of terms in the expansion is 28. So, we set up the equation: $\binom{n+2}{2} = 28$

• Solving for 'n': Recall that $\binom{N}{2} = \frac{N(N-1)}{2}$. Applying this to our equation with $N = n+2$: $\frac{(n+2)( (n+2) - 1 )}{2} = 28$ $\frac{(n+2)(n+1)}{2} = 28$ Multiply both sides by 2: $(n+2)(n+1) = 56$ Now, we need to find two consecutive integers whose product is 56. We know that $7 \times 8 = 56$. Comparing $(n+1)(n+2)$ with $7 \times 8$, we can see that $n+1 = 7$ (and thus $n+2=8$). Therefore, $n+1 = 7$ $n = 6$ Alternatively, we can expand the equation to form a quadratic: $n^2 + 3n + 2 = 56$ $n^2 + 3n - 54 = 0$ Factoring the quadratic equation: $(n+9)(n-6) = 0$ This gives two possible values for $n$: $n = -9$ or $n = 6$. Since $n$ must be a positive integer for a valid expansion power, we discard $n=-9$. Thus, the value of $n$ is $\mathbf{6}$.

Step 2: Calculate the sum of the coefficients

• Principle: To find the sum of the coefficients of all terms in the expansion of ${\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}$, we substitute $x=1$ into the base expression.

• Substitution and Calculation: Substitute $x=1$ into the base of the given expression: $\left( 1 - \frac{2}{1} + \frac{4}{1^2} \right)^n$ $= (1 - 2 + 4)^n$ $= (3)^n$ Now, substitute the value of $n=6$ that we found in Step 1: $= (3)^6$ Calculate the value: $3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 = 81 \times 9 = 729$ Thus, the sum of the coefficients of all the terms in this expansion is $\mathbf{729}$.

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Tips and Common Mistakes

• Multinomial vs. Binomial: Always be careful to use the correct formula for the number of terms. For a binomial $(a+b)^n$, the number of terms is $n+1$. For a trinomial $(a+b+c)^n$, it's $\binom{n+2}{2}$. For a general multinomial with $k$ terms, it's $\binom{n+k-1}{k-1}$.
• Sum of Coefficients: This is a very common and useful trick. Remember that for any polynomial $P(x_1, x_2, \dots, x_k)$, the sum of its coefficients is $P(1, 1, \dots, 1)$.
• Algebraic Errors: Double-check your calculations when solving for $n$, especially when dealing with quadratic equations. Ensure you select the appropriate positive integer solution for $n$.

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Summary and Key Takeaway

This problem effectively tests two distinct but related concepts: determining the number of terms in a multinomial expansion and finding the sum of its coefficients. By correctly applying the formula for the number of terms ($\binom{n+k-1}{k-1}$) to find $n=6$, and then utilizing the property that substituting all variables with 1 yields the sum of coefficients, we found the sum to be $3^6 = 729$. This demonstrates the power of these general properties in simplifying complex expansion problems.