Key Concepts: Binomial Theorem and Geometric Progression

This problem requires the application of two core mathematical concepts:

1. Binomial Theorem: Specifically, the expansion of $(1-x)^n$, which is given by the sum: $(1-x)^n = \sum_{k=0}^n (-1)^k {}^n{C_k} x^k$ This identity is crucial for simplifying the initial complex summation.
2. Sum of a Geometric Progression (GP): For a finite geometric series with the first term $a$, a common ratio $r$, and $N$ terms, the sum $S_N$ is: $S_N = \frac{a(1-r^N)}{1-r}$ This formula will be used to sum the terms obtained after applying the Binomial Theorem.

Step-by-Step Solution

1. Deconstructing the Given Expression A The given expression for $A$ is: $A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]}$ The summation operator is linear, meaning we can distribute it across the terms inside the square brackets. This allows us to break down the complex sum into a sum of five simpler terms: $A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{1 \over 2}} \right)}^k}} + \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{3 \over 4}} \right)}^k}} + \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{7 \over 8}} \right)}^k}} + \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{{15} \over {16}}} \right)}^k}} + \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{{31} \over {32}}} \right)}^k}}$

2. Applying the Binomial Theorem to Each Individual Sum Each of these five sums perfectly matches the form of the binomial expansion $(1-x)^n = \sum_{k=0}^n (-1)^k {}^n{C_k} x^k$. We apply this identity to each term:

• First term (where $x = \frac{1}{2}$): $\sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{1 \over 2}} \right)}^k}} = {\left( {1 - {1 \over 2}} \right)^n} = {\left( {{1 \over 2}} \right)^n}$ Explanation: By recognizing the binomial pattern, we simplify the entire summation into a single power term.

• Second term (where $x = \frac{3}{4}$): $\sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{3 \over 4}} \right)}^k}} = {\left( {1 - {3 \over 4}} \right)^n} = {\left( {{1 \over 4}} \right)^n} = {\left( {{1 \over 2^2}} \right)^n} = {\left( {{1 \over 2}} \right)^{2n}}$ Explanation: We simplify the base of the power $(1/4)$ to $(1/2)^2$ to reveal a consistent pattern with the first term, $(1/2)$, which will be useful for identifying a geometric progression later.

• Third term (where $x = \frac{7}{8}$): $\sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{7 \over 8}} \right)}^k}} = {\left( {1 - {7 \over 8}} \right)^n} = {\left( {{1 \over 8}} \right)^n} = {\left( {{1 \over 2^3}} \right)^n} = {\left( {{1 \over 2}} \right)^{3n}}$ Explanation: Following the same logic, we express $(1/8)$ as $(1/2)^3$.

• Fourth term (where $x = \frac{15}{16}$): $\sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{{15} \over {16}}} \right)}^k}} = {\left( {1 - {{15} \over {16}}} \right)^n} = {\left( {{1 \over {16}}} \right)^n} = {\left( {{1 \over 2^4}} \right)^n} = {\left( {{1 \over 2}} \right)^{4n}}$ Explanation: The pattern continues, with $(1/16)$ becoming $(1/2)^4$.

• Fifth term (where $x = \frac{31}{32}$): $\sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}{{\left( {{{31} \over {32}}} \right)}^k}} = {\left( {1 - {{31} \over {32}}} \right)^n} = {\left( {{1 \over {32}}} \right)^n} = {\left( {{1 \over 2^5}} \right)^n} = {\left( {{1 \over 2}} \right)^{5n}}$ Explanation: Finally, $(1/32)$ is rewritten as $(1/2)^5$.

3. Summing the Simplified Terms as a Geometric Progression Now, substitute these simplified terms back into the expression for $A$: $A = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + {\left( {{1 \over 2}} \right)^{4n}} + {\left( {{1 \over 2}} \right)^{5n}}$ This is clearly a finite Geometric Progression (GP) with:

• First term, $a = \left(\frac{1}{2}\right)^n$
• Common ratio, $r = \frac{\left(\frac{1}{2}\right)^{2n}}{\left(\frac{1}{2}\right)^n} = \left(\frac{1}{2}\right)^n$
• Number of terms, $N = 5$

Using the formula for the sum of a GP, $S_N = \frac{a(1-r^N)}{1-r}$: $A = \frac{{{{\left( {{1 \over 2}} \right)}^n}\left( {1 - {{\left( {{{\left( {{1 \over 2}} \right)}^n}} \right)}^5}} \right)}}{{1 - {{\left( {{1 \over 2}} \right)}^n}}}$ $A = \frac{{{{\left( {{1 \over 2}} \right)}^n}\left( {1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \right)}}{{1 - {{\left( {{1 \over 2}} \right)}^n}}}$ To simplify this complex fraction, we can rewrite it using powers of 2: $A = \frac{{\frac{1}{{{2^n}}}\left( {1 - \frac{1}{{{2^{5n}}}}} \right)}}{{1 - \frac{1}{{{2^n}}}}} = \frac{{\frac{1}{{{2^n}}}\left( {\frac{{{2^{5n}} - 1}}{{{2^{5n}}}}} \right)}}{{\frac{{{2^n} - 1}}{{{2^n}}}}}$ $A = \frac{{{2^{5n}} - 1}}{{{2^{5n}}({2^n} - 1)}}$ Explanation: This algebraic manipulation clears the fractional powers and combines the terms into a single, more manageable fraction, which is essential for the next step of equating it to the given condition.

4. Using the Given Condition to Solve for n We are provided with the equation relating $A$ to a specific value: $63A = 1 - {1 \over {{2^{30}}}}$ Substitute the simplified expression for $A$: $63 \left( \frac{{{2^{5n}} - 1}}{{{2^{5n}}({2^n} - 1)}} \right) = 1 - {1 \over {{2^{30}}}}$ To make the comparison easier, we can rewrite the left-hand side: $\frac{{63}}{{{2^n} - 1}} \cdot \frac{{{2^{5n}} - 1}}{{{2^{5n}}}} = \frac{{63}}{{{2^n} - 1}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$ So the equation becomes: $\frac{{63}}{{{2^n} - 1}}\left( {1 - {1 \over {{2^{5n}}}}} \right) = 1 - {1 \over {{2^{30}}}}$ Explanation: This step is a strategic algebraic rearrangement. By factoring out the term $\frac{63}{2^n - 1}$, the left side takes a form that is directly comparable to the right side, making it easier to identify the value of $n$ by inspection.

By comparing the two sides of the equation, we can infer the values that must hold true for equality:

1. The denominator of the fraction on the left must equal $63$: ${2^n} - 1 = 63$
2. The term in the parenthesis on the left must equal the entire right side of the given condition: $1 - {1 \over {{2^{5n}}}} = 1 - {1 \over {{2^{30}}}}$

Let's solve the first equation: ${2^n} - 1 = 63$ ${2^n} = 64$ Since $64 = 2^6$, we have: ${2^n} = {2^6} \implies n = 6$

Now, let's verify if this value of $n=6$ satisfies the second condition: Substitute $n=6$ into the second part: $1 - {1 \over {{2^{5 \times 6}}}} = 1 - {1 \over {{2^{30}}}}$ $1 - {1 \over {{2^{30}}}} = 1 - {1 \over {{2^{30}}}}$ Both conditions are consistently satisfied by $n=6$.

Therefore, the value of $n$ is 6.

Tips for Solving Similar Problems & Common Mistakes:

• Pattern Recognition is Key: Always look for known mathematical patterns (like binomial expansions, GP, AP, HP) when faced with complex sums or series.
• Simplify Bases to a Common Form: When dealing with fractions raised to powers (e.g., $(1/4)^n$, $(1/8)^n$), converting them to a common base (like $(1/2)^{2n}$, $(1/2)^{3n}$) is crucial for identifying underlying sequences like GP.
• Strategic Algebraic Manipulation: Don't be afraid to rearrange expressions. Sometimes, a different form of the same expression can reveal a solution much more easily, as seen when we rewrote $A$ to match the form of the given condition.
• Verification: After finding a solution (like $n=6$), always substitute it back into the original equations or conditions to ensure consistency and correctness.
• Avoid Calculation Errors: Problems like these, while conceptually straightforward, can be prone to minor arithmetic or exponent errors. Double-check your calculations.

Summary and Key Takeaway: This problem serves as an excellent example of how combining different mathematical concepts can lead to an elegant solution. The ability to recognize a binomial expansion hidden within a larger sum, and then to identify the resulting terms as a geometric progression, was fundamental. Finally, strategic algebraic manipulation and careful comparison were used to efficiently solve for the unknown variable $n$. This problem emphasizes the importance of a strong foundational understanding of series and sequences, along with keen observation skills.