Sequences & Series
Complex Sum — Set Cardinality
nta_pyq_2024_apr
Grade 11
Question:
If the set $R=\{(a,b):a+5b=42,\,a,b\in\mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^{m}(1-i^{n!})=x+iy$, where $i=\sqrt{-1}$, then the value of $m+x+y$ is
Step-by-Step Solution
Key Concept: $a+5b=42$, $a,b\in\mathbb{N}$: $b=1,2,\ldots,8$ (for $a=37,32,\ldots,2$). So $m=8$. $\sum_{n=1}^{8}(1-i^{n!})$: for $n\geq2$, $n!$ is divisible by 4 so $i^{n!}=1$. For $n=1$: $1-i^1=1-i$. Terms $n=2$ to $8$: $7\times(1-1)=0$.
$m=8$. Sum: $(1-i)+2+2+5\cdot0=5-i$. $x=5$, $y=-1$. $m+x+y=12$.
Correct Answer: 1