Question

Let (1 + x + 2x²)²⁰ = a₀ + a₁x + a₂x² + ⋯ + a₄₀x⁴⁰. Then, a₁ + a₃ + a₅ + ⋯ + a₃₇ is equal to :

• A. 2¹⁹(2²⁰ + 21)
• B. 2²⁰(2²⁰ + 21)
• C. 2¹⁹(2²⁰ - 21)
• D. 2²⁰(2²⁰ - 21)

Hint:

To find the sum of even/odd coefficients, evaluate the polynomial at $1$ and $-1$.

Answer 1

The Correct Option is C

Step 1: Let $P(x) = (1+x+2x^2)^{20}$.
Step 2: Sum of odd coefficients $S_{odd} = \frac{P(1) - P(-1)}{2}$.
Step 3: $P(1) = (1+1+2)^{20} = 4^{20} = 2^{40}$.
Step 4: $P(-1) = (1-1+2)^{20} = 2^{20}$.
Step 5: $a_1 + a_3 + ....... + a_{39} = \frac{2^{40} - 2^{20}}{2} = 2^{39} - 2^{19} = 2^{19}(2^{20} - 1)$.
Step 6: The question asks for sum up to $a_{37}$. We need to subtract $a_{39}$. $a_{39}$ is the coefficient of $x^{39}$. In $(1+x+2x^2)^{20}$, the highest power is $x^{40}$ (from $(2x^2)^{20}$). To get $x^{39}$, we must have nineteen $(2x^2)$ and one $(x)$. Coeff $a_{39} = \binom{20}{19, 1, 0} (2)^{19} (1)^1 = 20 \times 2^{19}$.
Step 7: Sum $= 2^{19}(2^{20}-1) - 20 \cdot 2^{19} = 2^{19}(2^{20} - 1 - 20) = 2^{19}(2^{20} - 21)$.