Solution: Sum of Odd-Indexed Coefficients in a Polynomial Expansion

Key Concept: Sum of Coefficients in a Polynomial Expansion

For a polynomial $P(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$, the sum of its coefficients can be found by substituting $x=1$ into the polynomial. $P(1) = a_0 + a_1 + a_2 + ... + a_n$ Similarly, the alternating sum of coefficients can be found by substituting $x=-1$: $P(-1) = a_0 - a_1 + a_2 - ... + (-1)^n a_n$ These two sums are crucial for isolating the sum of odd-indexed or even-indexed coefficients. To find the sum of odd-indexed coefficients ($a_1 + a_3 + a_5 + ...$), we use the formula: $a_1 + a_3 + a_5 + ... = \frac{P(1) - P(-1)}{2}$ And for the sum of even-indexed coefficients ($a_0 + a_2 + a_4 + ...$): $a_0 + a_2 + a_4 + ... = \frac{P(1) + P(-1)}{2}$

Step-by-Step Derivation

Given the expansion: $(1 + x + 2x^2)^{20} = a_0 + a_1x + a_2x^2 + .... + a_{40}x^{40}$ Let $P(x) = (1 + x + 2x^2)^{20}$.

Step 1: Calculate the sum of all coefficients by substituting $x=1$. We substitute $x=1$ into the polynomial $P(x)$ to obtain the sum of all coefficients, $a_0 + a_1 + ... + a_{40}$. $P(1) = (1 + 1 + 2(1)^2)^{20}$ $P(1) = (1 + 1 + 2)^{20}$ $P(1) = (4)^{20}$ So, we have our first equation: ${4^{20}} = {a_0} + {a_1} + {a_2} + ....... + {a_{40}}$ Let's call this Equation (i).

Step 2: Calculate the alternating sum of coefficients by substituting $x=-1$. Next, we substitute $x=-1$ into $P(x)$ to find the alternating sum of coefficients. This is done because powers of $x$ become alternating in sign when $x=-1$, allowing us to later isolate either odd or even terms. $P(-1) = (1 + (-1) + 2(-1)^2)^{20}$ $P(-1) = (1 - 1 + 2(1))^{20}$ $P(-1) = (2)^{20}$ This gives us our second equation: ${2^{20}} = {a_0} - {a_1} + {a_2} - ....... - {a_{39}} + {a_{40}}$ Let's call this Equation (ii).

Step 3: Combine equations to find the sum of odd-indexed coefficients. To find the sum of odd-indexed coefficients ($a_1 + a_3 + ...$), we subtract Equation (ii) from Equation (i). This operation cancels out all the even-indexed coefficients ($a_0, a_2, ...$) and doubles the odd-indexed coefficients ($a_1, a_3, ...$). $(i) - (ii):$ ${4^{20}} - {2^{20}} = ({a_0} + {a_1} + ... + {a_{40}}) - ({a_0} - {a_1} + ... - {a_{39}} + {a_{40}})$ ${4^{20}} - {2^{20}} = 2({a_1} + {a_3} + ...... + {a_{39}})$ Now, we can solve for the sum of odd-indexed coefficients up to $a_{39}$: ${a_1} + {a_3} + ...... + {a_{39}} = \frac{{4^{20}} - {2^{20}}}{2}$ We simplify the expression using exponent rules, noting that $4^{20} = (2^2)^{20} = 2^{40}$: ${a_1} + {a_3} + ...... + {a_{39}} = \frac{{2^{40}} - {2^{20}}}{2}$ ${a_1} + {a_3} + ...... + {a_{39}} = \frac{{2^{40}}}{2} - \frac{{2^{20}}}{2}$ ${a_1} + {a_3} + ...... + {a_{39}} = {2^{39}} - {2^{19}}$ Let's call this Equation (iii).

Step 4: Adjust the sum for the required range. The question asks for the sum $a_1 + a_3 + ... + a_{37}$. Our current sum from Equation (iii) includes $a_{39}$. Therefore, to get the desired sum, we need to subtract $a_{39}$ from the result of Equation (iii): ${a_1} + {a_3} + ...... + {a_{37}} = ({2^{39}} - {2^{19}}) - {a_{39}}$ To proceed, we must calculate the value of $a_{39}$.

Step 5: Calculate the coefficient $a_{39}$ using the Multinomial Theorem. The term $a_{39}$ is the coefficient of $x^{39}$ in the expansion of $(1 + x + 2x^2)^{20}$. According to the Multinomial Theorem, the general term in the expansion of $(y_1 + y_2 + y_3)^n$ is given by: $\frac{n!}{p!q!r!} (y_1)^p (y_2)^q (y_3)^r$ where $p+q+r = n$. In our case, $y_1=1$, $y_2=x$, $y_3=2x^2$, and $n=20$. So, the general term is: $\frac{20!}{p!q!r!} (1)^p (x)^q (2x^2)^r = \frac{20!}{p!q!r!} (2)^r x^{q+2r}$ We need the coefficient of $x^{39}$, which means we need to find non-negative integer values for $p, q, r$ such that:

1. $p+q+r = 20$ (sum of powers must equal the total power of the expansion)
2. $q+2r = 39$ (sum of powers of $x$ must equal 39)

Let's analyze the second condition, $q+2r=39$: Since $r$ is a non-negative integer and $q \ge 0$, the maximum possible value for $r$ occurs when $q=0$. This would mean $2r=39$, which is not possible for integer $r$. The closest even number to 39 is 38. If $2r=38$, then $r=19$. If $r=19$: $q + 2(19) = 39$ $q + 38 = 39$ $q = 1$ Now check the first condition with $r=19$ and $q=1$: $p+1+19 = 20$ $p+20 = 20$ $p = 0$ This gives us a unique set of values: $p=0, q=1, r=19$. Now substitute these values into the general term formula to find $a_{39}$: $a_{39} = \frac{20!}{0!1!19!} (1)^0 (1)^1 (2)^{19}$ $a_{39} = \frac{20 \times 19!}{1 \times 1 \times 19!} \times 1 \times 1 \times 2^{19}$ $a_{39} = 20 \times 2^{19}$

Step 6: Final Calculation of the required sum. Now we substitute the value of $a_{39}$ back into the expression from Step 4: ${a_1} + {a_3} + ...... + {a_{37}} = ({2^{39}} - {2^{19}}) - (20 \times {2^{19}})$ Factor out the common term $2^{19}$: ${a_1} + {a_3} + ...... + {a_{37}} = {2^{19}} (2^{20} - 1) - (20 \times {2^{19}})$ ${a_1} + {a_3} + ...... + {a_{37}} = {2^{19}} (2^{20} - 1 - 20)$ ${a_1} + {a_3} + ...... + {a_{37}} = {2^{19}} (2^{20} - 21)$

Comparing this with the given options, the correct answer is (B). However, the original problem states (A) as the correct answer. Let's recheck the options. Option (A) is $2^{20}(2^{20} - 21)$. Option (B) is $2^{19}(2^{20} - 21)$.

My derived answer is $2^{19}(2^{20} - 21)$. Let's double check if I misread the target answer or if there's a miscalculation. The problem statement gives "Correct Answer: A". My calculation yields $2^{19}(2^{20} - 21)$. This means there might be a discrepancy between the provided solution's final simplification and the option choice. Let's re-examine the given solution's final steps.

Given solution's final steps: $\Rightarrow {a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}} - {a_{39}}$ ..... (iii) (This is effectively my step 4) $a_{39} = 20 \times 2^{19}$ (Matches my step 5) $\therefore$ ${a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}}.21$ (Here, they combine $-2^{19}$ and $-a_{39}$ as $-2^{19} - 20 \cdot 2^{19} = -2^{19}(1+20) = -21 \cdot 2^{19}$. This is correct.) $\Rightarrow {2^{19}}({2^{20}} - 21)$ (This factorization is also correct: $2^{39} - 21 \cdot 2^{19} = 2^{19}(2^{20} - 21)$).

The calculated answer $2^{19}(2^{20} - 21)$ indeed matches Option (B), not (A). If the intended answer was Option (A) $2^{20}(2^{20} - 21)$, then my calculations, and the provided solution's calculation, are consistently leading to Option (B). There might be a typo in the provided "Correct Answer: A".

Assuming my calculation is correct and the provided Correct Answer: A is a typo, my final answer is $2^{19}(2^{20} - 21)$.

Tips for Success / Common Pitfalls

1. Careful with Indices: Pay close attention to the highest index in the sum requested ($a_{37}$ vs $a_{39}$). This often requires calculating the last term separately.
2. Multinomial Theorem Precision: When using the multinomial theorem, ensure that both conditions (sum of powers of terms equals total power, and sum of powers of $x$ equals the required exponent) are satisfied. Also, correctly calculate factorials and powers.
3. Exponent Rules: Be meticulous with exponent simplification, especially with terms like $4^{20} = (2^2)^{20} = 2^{40}$.
4. Algebraic Manipulation: Factorization can simplify the final expression. Look for common factors like $2^{19}$.

Summary / Key Takeaway

This problem effectively tests two important concepts in Binomial Theorem:

1. Utilizing $P(1)$ and $P(-1)$: A standard technique to find sums of odd or even indexed coefficients in a polynomial expansion.
2. Multinomial Theorem for specific coefficients: Applying the multinomial theorem to find the coefficient of a particular term in an expansion with three or more terms. The combination of these two techniques is essential for solving such problems efficiently. Always double-check the exact range of the sum required to avoid off-by-one errors in the terms.