Key Concepts and Formulas

This problem involves concepts from the Binomial Theorem, specifically:

1. Sum of coefficients of odd/even powers of $x$: For a binomial expansion $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \dots + C_n x^n$,
   • The sum of all coefficients is $C_0 + C_1 + C_2 + \dots + C_n = (1+1)^n = 2^n$.
   • The sum of coefficients of even powers ($C_0 + C_2 + C_4 + \dots$) is $2^{n-1}$.
   • The sum of coefficients of odd powers ($C_1 + C_3 + C_5 + \dots$) is $2^{n-1}$. This property is derived by considering $(1+1)^n$ and $(1-1)^n$.
2. General Term in Binomial Expansion: The $(r+1)^{th}$ term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_r a^{n-r} b^r$.
3. Middle Term in Binomial Expansion: If $n$ is an even number, say $n=2k$, then the expansion of $(a+b)^{2k}$ has $(2k+1)$ terms, and there is only one middle term, which is the $(k+1)^{th}$ term. Its value is $T_{k+1} = {}^{2k}C_k a^k b^k$.
4. Ratio of Binomial Coefficients: A useful identity for simplifying ratios of consecutive binomial coefficients is ${}^{n}C_r / {}^{n}C_{r+1} = \frac{r+1}{n-r}$.

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Step 1: Finding K (Sum of coefficients of odd powers of $x$ in $(1+x)^{99})$

• Concept Applied: Sum of coefficients of odd powers in $(1+x)^n$.
• Explanation: We need to find the sum of coefficients of odd powers of $x$ in the expansion of $(1+x)^{99}$. According to the property mentioned above, for $(1+x)^n$, the sum of coefficients of odd powers of $x$ is $2^{n-1}$.
• Calculation: Here, $n = 99$. So, $K = 2^{99-1} = 2^{98}$.
• Why this step: This directly applies a standard binomial theorem property to efficiently find K without expanding the entire expression.

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Step 2: Finding $a$ (Middle term in the expansion of ${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}})$

• Concept Applied: Middle term in a binomial expansion with an even power.
• Explanation: The given binomial is ${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$. The power is $n = 200$, which is an even number. When $n$ is even, there is a single middle term, which is the $(\frac{n}{2} + 1)^{th}$ term. So, the middle term is the $(\frac{200}{2} + 1)^{th} = (100+1)^{th} = 101^{st}$ term. Using the general term formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$, for the $101^{st}$ term, we have $r=100$, $n=200$, first term is $2$, and second term is ${1 \over {\sqrt 2 }}$.
• Calculation: $a = T_{101} = {}^{200}C_{100} (2)^{200-100} \left({1 \over {\sqrt 2 }}\right)^{100}$ $a = {}^{200}C_{100} (2)^{100} \left( (2^{-1/2})^{100} \right)$ $a = {}^{200}C_{100} (2)^{100} (2^{-50})$ $a = {}^{200}C_{100} (2)^{100-50}$ $a = {}^{200}C_{100} (2)^{50}$
• Why this step: Correctly identifying and calculating the middle term is crucial for the overall expression. Careful application of exponent rules is important here.

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Step 3: Evaluating the Expression ${{{}^{200}{C_{99}}K} \over a}$ and expressing it as ${{{2^l}m} \over n}$

• Concept Applied: Substitution, simplification of binomial coefficients, and exponent rules.
• Explanation: Now we substitute the values of $K$ and $a$ we found into the given expression and simplify it. We then need to manipulate the result to match the form ${{{2^l}m} \over n}$, where $m$ and $n$ are odd integers.
• Calculation: Substitute $K = 2^{98}$ and $a = {}^{200}C_{100} 2^{50}$: $E = {{{}^{200}{C_{99}}K} \over a} = {{{}^{200}{C_{99}} \cdot {2^{98}}} \over {{}^{200}{C_{100}} \cdot {2^{50}}}}$ Separate the binomial coefficients and the powers of 2: $E = \left( {{{}^{200}{C_{99}}} \over {{}^{200}{C_{100}}}} \right) \cdot \left( {{{2^{98}}} \over {{2^{50}}}} \right)$ Simplify the ratio of binomial coefficients: ${{{}^{200}{C_{99}}} \over {{}^{200}{C_{100}}}} = \frac{\frac{200!}{99!(200-99)!}}{\frac{200!}{100!(200-100)!}} = \frac{200!}{99!101!} \cdot \frac{100!100!}{200!} = \frac{100!100!}{99!101!} = \frac{(100 \cdot 99!) \cdot 100!}{99! \cdot (101 \cdot 100!)} = \frac{100}{101}$ Simplify the powers of 2: ${{{2^{98}}} \over {{2^{50}}}} = 2^{98-50} = 2^{48}$ Combine the simplified parts: $E = \frac{100}{101} \cdot 2^{48}$ To match the form ${{{2^l}m} \over n}$ where $m$ and $n$ are odd, we need to factor out all powers of 2 from the numerator and denominator. We know $100 = 4 \cdot 25 = 2^2 \cdot 25$. $E = \frac{2^2 \cdot 25}{101} \cdot 2^{48}$ Combine the powers of 2 in the numerator: $E = \frac{2^{2+48} \cdot 25}{101} = \frac{2^{50} \cdot 25}{101}$ Now, compare this with ${{{2^l}m} \over n}$: $l = 50$ $m = 25$ (which is an odd number) $n = 101$ (which is an odd number)
• Why this step: This is the core calculation of the problem. It requires careful handling of factorials and exponents. The final manipulation to ensure $m$ and $n$ are odd is a critical requirement of the problem statement.

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Step 4: Determining the ordered pair $(l, n)$

• Concept Applied: Direct extraction from the simplified form.
• Explanation: Based on our simplification in Step 3, we have successfully expressed the given quantity in the required format. Now we just need to identify the values of $l$ and $n$.
• Result: From the expression ${{{2^{50}} \cdot 25} \over {101}}$, we have $l = 50$ and $n = 101$. Therefore, the ordered pair $(l, n)$ is $(50, 101)$.
• Why this step: This provides the final answer to the question asked.

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Common Mistakes and Tips

• Binomial Coefficient Ratio: Be careful with the formula for ${}^{n}C_r / {}^{n}C_{r+1}$. It's easy to confuse the numerator and denominator. Always re-derive or double-check if unsure.
• Middle Term: For even powers $(2k)$, there is one middle term, which is the $(k+1)^{th}$. For odd powers $(2k+1)$, there are two middle terms, the $(k+1)^{th}$ and $(k+2)^{th}$.
• Powers of Two: When simplifying to the form ${{{2^l}m} \over n}$, ensure that $m$ and $n$ are indeed odd. This means factoring out all possible powers of 2 into $2^l$. For example, if you had $\frac{2^3 \cdot 6}{5}$, you would write it as $\frac{2^3 \cdot (2 \cdot 3)}{5} = \frac{2^4 \cdot 3}{5}$, so $m=3$ and $n=5$.

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Summary and Key Takeaway

This problem effectively tests understanding of fundamental binomial theorem properties: calculating sums of coefficients for specific power types, finding the middle term, and simplifying expressions involving binomial coefficients and exponents. The key takeaway is to apply each property systematically and simplify carefully, paying close attention to the form required for the final answer, especially the odd/even nature of $m$ and $n$. The final ordered pair $(l,n)$ is $(50, 101)$.