Key Concepts

This problem primarily utilizes two fundamental concepts from the Binomial Theorem:

1. Sum of Odd Coefficients: For a polynomial expansion $f(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$, the sum of the odd coefficients ($a_1 + a_3 + a_5 + \ldots$) can be found using the formula: $S_{\text{odd}} = a_1 + a_3 + a_5 + \ldots = \frac{f(1) - f(-1)}{2}$ Why this works: When $x=1$, $f(1) = a_0 + a_1 + a_2 + a_3 + \ldots + a_n$. When $x=-1$, $f(-1) = a_0 - a_1 + a_2 - a_3 + \ldots + (-1)^n a_n$. Subtracting the second equation from the first gives: $f(1) - f(-1) = (a_0 + a_1 + a_2 + \ldots) - (a_0 - a_1 + a_2 - \ldots) = 2(a_1 + a_3 + a_5 + \ldots) = 2 S_{\text{odd}}$. Hence, $S_{\text{odd}} = \frac{f(1) - f(-1)}{2}$.

2. Coefficient of a Specific Term: To find the coefficient of a specific term $x^r$ in a more complex expansion, we can often manipulate the expression into a product of simpler series, or use the generalized binomial theorem for negative or fractional exponents. The generalized binomial theorem states that for any real number $\alpha$, $(1+y)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} y^n = 1 + \alpha y + \frac{\alpha(\alpha-1)}{2!}y^2 + \ldots$ A common specific case is for $(1-y)^{-n}$, where the coefficient of $y^r$ is $\binom{n+r-1}{r}$ or $\binom{n+r-1}{n-1}$. For $(1-y)^{-n} = \sum_{r=0}^{\infty} \binom{n+r-1}{r} y^r$, which can also be written as $\binom{n+r-1}{n-1}$.

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Step-by-Step Solution

1. Define the given expansion Let the given expansion be $f(x) = \left(1+x+x^2\right)^{10}$. We are given that this expansion is equal to $\sum_{r=0}^{20} a_r x^r = a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$. Here, $a_r$ represents the coefficient of $x^r$ in the expansion of $f(x)$.

2. Calculate the sum of odd coefficients ($a_1+a_3+a_5+\ldots+a_{19}$)

We use the formula for the sum of odd coefficients: $S_{\text{odd}} = \frac{f(1) - f(-1)}{2}$.

• Step 2a: Evaluate $f(1)$ Substitute $x=1$ into the expression for $f(x)$: $f(1) = (1+1+1)^{10} = (3)^{10}$ We calculate $3^{10}$: $3^1 = 3$ $3^2 = 9$ $3^3 = 27$ $3^4 = 81$ $3^5 = 243$ $3^{10} = (3^5)^2 = (243)^2 = 59049$. So, $f(1) = 59049$.

• Step 2b: Evaluate $f(-1)$ Substitute $x=-1$ into the expression for $f(x)$: $f(-1) = (1+(-1)+(-1)^2)^{10} = (1-1+1)^{10} = (1)^{10} = 1$ So, $f(-1) = 1$.

• Step 2c: Calculate $S_{\text{odd}}$ Now, substitute the values of $f(1)$ and $f(-1)$ into the formula: $S_{\text{odd}} = \frac{59049 - 1}{2} = \frac{59048}{2} = 29524$ Thus, $a_1+a_3+a_5+\ldots+a_{19} = 29524$.

3. Calculate the coefficient $a_2$

To find $a_2$, the coefficient of $x^2$, we can rewrite the expression $1+x+x^2$ using the sum of a geometric series: $1+x+x^2 = \frac{1-x^3}{1-x}$ Why this transformation: This transformation is useful because it allows us to express $f(x)$ as a product of two terms, $(1-x^3)^{10}$ and $(1-x)^{-10}$, which can be expanded using standard binomial theorems.

So, $f(x) = \left(\frac{1-x^3}{1-x}\right)^{10} = (1-x^3)^{10} (1-x)^{-10}$.

Now, we need to find the coefficient of $x^2$ in the product of these two expansions:

• Step 3a: Expand $(1-x^3)^{10}$ Using the binomial theorem $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$: $(1-x^3)^{10} = \binom{10}{0}(1)^{10}(-x^3)^0 + \binom{10}{1}(1)^9(-x^3)^1 + \binom{10}{2}(1)^8(-x^3)^2 + \ldots$ $(1-x^3)^{10} = 1 - 10x^3 + 45x^6 - \ldots$ Why we look at these terms: We are interested in finding the coefficient of $x^2$ in the overall product. The term $x^3$ from this expansion already has a power greater than $2$. Therefore, only the constant term ($x^0$) from this expansion can contribute to the $x^2$ term when multiplied with terms from the other expansion. The coefficient of $x^0$ in $(1-x^3)^{10}$ is $\binom{10}{0} = 1$.

• Step 3b: Expand $(1-x)^{-10}$ Using the generalized binomial theorem for $(1-y)^{-n}$, where the coefficient of $y^r$ is $\binom{n+r-1}{r}$: Here, $y=x$ and $n=10$. We need the coefficient of $x^2$ from this expansion to combine with the $x^0$ term from $(1-x^3)^{10}$. The coefficient of $x^r$ in $(1-x)^{-10}$ is $\binom{10+r-1}{r} = \binom{9+r}{r}$. For $x^2$ (i.e., $r=2$), the coefficient is: $\binom{9+2}{2} = \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55$ Alternatively, using $\binom{n+r-1}{n-1}$: $\binom{10+2-1}{10-1} = \binom{11}{9} = \frac{11 \times 10}{2 \times 1} = 55$

• Step 3c: Combine to find $a_2$ The term $x^2$ in the product $f(x) = (1-x^3)^{10} (1-x)^{-10}$ is obtained by multiplying: (coefficient of $x^0$ in $(1-x^3)^{10}$) $\times$ (coefficient of $x^2$ in $(1-x)^{-10}$) This is because any $x^{3k}$ term from $(1-x^3)^{10}$ where $k \ge 1$ will result in a power of $x$ greater than or equal to $3$, so it cannot contribute to $x^2$. Therefore, $a_2 = (1) \times (55) = 55$.

  Alternative method for $a_2$ using Multinomial Theorem (for understanding): The general term in the expansion of $(1+x+x^2)^{10}$ is $\frac{10!}{p!q!r!} (1)^p (x)^q (x^2)^r$ where $p+q+r=10$. For the coefficient of $x^2$, we need $q+2r=2$. Possible pairs of $(q, r)$ that satisfy this are:

  1. If $r=0$, then $q=2$. In this case, $p=10-2-0=8$. The coefficient is $\frac{10!}{8!2!0!} = \frac{10 \times 9}{2 \times 1} = 45$.
  2. If $r=1$, then $q=0$. In this case, $p=10-0-1=9$. The coefficient is $\frac{10!}{9!0!1!} = \frac{10}{1} = 10$. Summing these up: $a_2 = 45 + 10 = 55$. This confirms the result from the previous method.

4. Final Calculation for $k$

We are given the equation: $\left(a_1+a_3+a_5+\ldots+a_{19}\right)-11 a_2=121 k$. Substitute the values we found for $S_{\text{odd}}$ and $a_2$: $29524 - 11 \times 55 = 121 k$ $29524 - 605 = 121 k$ $28919 = 121 k$ Why we isolate $k$: The problem asks for the value of $k$.

Now, solve for $k$: $k = \frac{28919}{121}$ To perform the division, we can do long division or recognize that $121 = 11^2$. $28919 \div 121$: $28919 / 11 = 2629$ $2629 / 11 = 239$ So, $k = 239$.

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Tips and Common Mistakes to Avoid

• Calculation Errors: Powers of $3$ (especially $3^{10}$) and multiplication/division can be prone to error. Double-check your arithmetic.
• Forgetting $(-1)^k$: When expanding $(1-x^3)^{10}$ or similar terms, be careful with the signs introduced by negative terms.
• Incorrect Binomial Coefficients: Ensure you use the correct formula for generalized binomial coefficients $\binom{n+r-1}{r}$ for negative powers like $(1-x)^{-n}$. A common mistake is to confuse it with $\binom{n}{r}$.
• Missing Terms in Coefficient Calculation: When combining expansions, make sure to account for all combinations of terms that multiply to the desired power of $x$. For $a_2$, only $x^0$ from the first series and $x^2$ from the second were relevant. For higher powers, multiple combinations might exist (e.g., $x^1 \cdot x^{r-1}$, $x^2 \cdot x^{r-2}$, etc.).
• Understanding the Formulas: Don't just memorize the formulas for sum of odd/even coefficients; understand their derivation. This helps in recall and applying them correctly.

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Summary and Key Takeaway

This problem effectively tests the application of two core principles of binomial expansions: calculating the sum of specific coefficients and finding the coefficient of a particular term. The strategy involved rewriting the expression to leverage standard binomial series and then carefully combining the terms. The ability to recognize and apply the identity $1+x+x^2 = \frac{1-x^3}{1-x}$ was crucial for simplifying the coefficient calculation. Always break down complex problems into smaller, manageable parts and re-verify calculations to avoid errors.