Key Concepts

This problem involves the Binomial Theorem and properties of binomial expansion, specifically the general term and the conditions for the power of a variable.

The Binomial Theorem states that for any non-negative integer $N$: $(a+b)^N = \sum_{r=0}^{N} {}^N C_r a^{N-r} b^r$ The general term of the expansion, denoted as $T_{r+1}$, is given by ${T_{r + 1}} = {}^N{C_r}\,{a^{N - r}}\,{b^r}$. In this problem, we are given the sum of the coefficients of positive powers of x. It is crucial to understand that "sum of coefficients" usually refers to substituting variables with 1. However, here we are only considering terms where the exponent of x is strictly positive.

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Step 1: Determine the General Term of the Binomial Expansion

Given the binomial expression is ${\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$. Here, $a = x^n$, $b = \frac{2}{x^5}$, and $N = 7$. Using the general term formula for binomial expansion: ${T_{r + 1}} = {}^7{C_r}\,{\left( {{x^n}} \right)^{7 - r}}\,{\left( {{2 \over {{x^5}}}} \right)^r}$ Now, we simplify this expression to separate the powers of x and the constant coefficient: ${T_{r + 1}} = {}^7{C_r}\,{x^{n(7 - r)}}\,{2^r}\,{\left( {{x^{ - 5}}} \right)^r}$ ${T_{r + 1}} = {}^7{C_r}\,{2^r}\,{x^{7n - nr - 5r}}$ ${T_{r + 1}} = {}^7{C_r}\,{2^r}\,{x^{7n - r(n + 5)}}$ The coefficient of the $(r+1)^{th}$ term is ${}^7{C_r}\,{2^r}$, and the exponent of $x$ is $7n - r(n + 5)$.

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Step 2: Establish the Condition for Positive Powers of x

We are interested in terms where the power of $x$ is positive. Therefore, the exponent of $x$ must satisfy: $7n - r(n + 5) > 0$ Rearranging this inequality to isolate $r$: $7n > r(n + 5)$ Since $n$ is a positive integer (it represents a power of $x$, usually $x^n$ where $n \ge 1$), $n+5$ will be positive. Thus, we can divide by $(n+5)$ without changing the direction of the inequality: $r < {{7n} \over {n + 5}}$ Since $r$ is the index of the term in the binomial expansion, $r$ must be an integer, and its possible values range from $0$ to $7$. So, $0 \le r \le 7$.

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Step 3: Calculate the Coefficients and Determine the Range of r

The problem states that the sum of the coefficients of all the positive powers of x is 939. The coefficient of the $(r+1)^{th}$ term is ${}^7{C_r}\,{2^r}$. We need to sum these coefficients for all $r$ that satisfy the condition $r < {{7n} \over {n + 5}}$ and $0 \le r \le 7$.

Let's calculate the value of ${}^7{C_r}\,{2^r}$ for each possible value of $r$:

• For $r = 0$: ${}^7{C_0}\,{2^0} = 1 \times 1 = 1$
• For $r = 1$: ${}^7{C_1}\,{2^1} = 7 \times 2 = 14$
• For $r = 2$: ${}^7{C_2}\,{2^2} = 21 \times 4 = 84$
• For $r = 3$: ${}^7{C_3}\,{2^3} = 35 \times 8 = 280$
• For $r = 4$: ${}^7{C_4}\,{2^4} = 35 \times 16 = 560$
• For $r = 5$: ${}^7{C_5}\,{2^5} = 21 \times 32 = 672$
• For $r = 6$: ${}^7{C_6}\,{2^6} = 7 \times 64 = 448$
• For $r = 7$: ${}^7{C_7}\,{2^7} = 1 \times 128 = 128$

Now, let's find the cumulative sum of these coefficients to match 939:

• Sum for $r=0$: $1$
• Sum for $r=0, 1$: $1 + 14 = 15$
• Sum for $r=0, 1, 2$: $15 + 84 = 99$
• Sum for $r=0, 1, 2, 3$: $99 + 280 = 379$
• Sum for $r=0, 1, 2, 3, 4$: $379 + 560 = 939$

We found that the sum of coefficients is 939 when terms up to $r=4$ are included. This implies that for $r = 0, 1, 2, 3, 4$, the power of $x$ must be positive. Conversely, for $r = 5, 6, 7$, the power of $x$ must be non-positive (either zero or negative), as these terms are not included in the sum of positive powers.

Therefore, the condition for $r$ must be $r \le 4$. Combining this with $r < {{7n} \over {n + 5}}$, it means that $r=4$ must satisfy the inequality, while $r=5$ must not.

This gives us the critical condition: $4 < {{7n} \over {n + 5}} \le 5$ Why this range?

• The terms up to $r=4$ have positive powers of $x$. This means $r=4$ must satisfy the condition $r < \frac{7n}{n+5}$, hence $\frac{7n}{n+5}$ must be strictly greater than $4$.
• The terms from $r=5$ onwards do not have positive powers of $x$. This means $r=5$ must not satisfy the condition $r < \frac{7n}{n+5}$. This implies $5 \ge \frac{7n}{n+5}$.

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Step 4: Solve the Inequalities for n

We need to solve the compound inequality: $4 < {{7n} \over {n + 5}} \le 5$

Part 1: $4 < {{7n} \over {n + 5}}$ Since $n$ is a positive integer, $n+5$ is always positive. We can multiply both sides by $(n+5)$ without changing the inequality direction. $4(n + 5) < 7n$ $4n + 20 < 7n$ $20 < 3n$ $n > {{20} \over 3}$ $n > 6.66...$

Part 2: ${{7n} \over {n + 5}} \le 5$ Again, multiply both sides by $(n+5)$: $7n \le 5(n + 5)$ $7n \le 5n + 25$ $2n \le 25$ $n \le {{25} \over 2}$ $n \le 12.5$

Combining the results: From Part 1, $n > 6.66...$ From Part 2, $n \le 12.5$ So, the combined range for $n$ is: $6.66... < n \le 12.5$

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Step 5: Find Possible Integral Values of n and Their Sum

Since $n$ must be an integer, the possible values for $n$ within the range $6.66... < n \le 12.5$ are: $n = 7, 8, 9, 10, 11, 12$

The sum of all possible integral values of $n$ is: Sum $= 7 + 8 + 9 + 10 + 11 + 12$ This is an arithmetic progression sum, or can be summed directly: Sum $= (7+12) + (8+11) + (9+10)$ Sum $= 19 + 19 + 19$ Sum $= 3 \times 19 = 57$

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Tips for Success & Common Mistakes

• Understanding the Question: Carefully read "sum of the coefficients of all the positive powers of x". This is not the total sum of coefficients (which would be found by setting $x=1$ in the expanded form) but a conditional sum.
• Indices vs. Exponents: Keep track of the 'r' from the general term formula and the exponent of 'x'. They are related but distinct concepts in the inequalities.
• Inequality Direction: When manipulating inequalities, remember that multiplying or dividing by a negative number reverses the inequality sign. In this case, $n+5$ is always positive for $n \ge 1$, so the direction remains unchanged.
• Strict vs. Non-Strict Inequality: The distinction between $r < K$ and $r \le K$ is critical. If $r=4$ is the last term with a positive power, then $4 < \frac{7n}{n+5}$. If $r=5$ is the first term with a non-positive power, then $5 \ge \frac{7n}{n+5}$. Combining these two gives the correct compound inequality $4 < \frac{7n}{n+5} \le 5$.

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Summary

By first determining the general term of the binomial expansion, we found the exponent of $x$ to be $7n - r(n+5)$ and the coefficient to be ${}^7{C_r}2^r$. We then calculated the cumulative sum of these coefficients until it matched the given value of 939, which indicated that terms up to $r=4$ had positive powers of $x$. This led to the critical inequality $4 < {{7n} \over {n + 5}} \le 5$. Solving this inequality for integer $n$ yielded values $7, 8, 9, 10, 11, 12$. The sum of these values is 57.