Binomial Theorem
Alternating binomial sum with powers of 3
MJAT_TS4_P2
Grade 12
Question:
The integers $a$ and $b$ satisfy $a^b = 3^0\binom{40}{0}-3^1\binom{40}{2}+3^2\binom{40}{4}-\cdots+(-1)^{20}\cdot3^{20}\binom{40}{40}$, where $b\in(10,15)$. Then $a+b=$
Step-by-Step Solution
Key Concept: The sum is $\text{Re}[(1+i\sqrt{3})^{40}]$. Write $1+i\sqrt{3}=2e^{i\pi/3}$, so $(1+i\sqrt{3})^{40}=2^{40}e^{i40\pi/3}$. $40\pi/3 = 13\pi+\pi/3$, so $\text{Re}[(1+i\sqrt{3})^{40}]=2^{40}\cos(40\pi/3)=2^{40}\cos(13\pi+\pi/3)=-2^{40}\cdot(1/2)=-2^{39}$.
$a^b=-2^{39}=(-8)^{13}$. $|a|+b=8+13=\mathbf{21}$.
Correct Answer: 21