Kerala KEAM 5 June 2024 Solved Paper

We are asked to evaluate the sum: $$\sum_{n=1}^{24} \left( i^n + i^{n+1} \right).$$Step 1: Use the periodicity of powers of $i$. The powers of $i$ repeat every four terms:$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad and then the cycle repeats.$$Thus, the powers of $i$ repeat every four terms.Step 2: Notice that $i^n + i^{n+1}$ will follow a periodic pattern as well. Let's compute the first few terms to identify the pattern:- For $n = 1$, $i^1 + i^2 = i + (-1) = i - 1$,- For $n = 2$, $i^2 + i^3 = -1 + (-i) = -1 - i$,- For $n = 3$, $i^3 + i^4 = -i + 1 = 1 - i$,- For $n = 4$, $i^4 + i^5 = 1 + i = 1 + i$.Step 3: The sum $i^n + i^{n+1}$ follows the pattern $i - 1, -1 - i, 1 - i, 1 + i$, which repeats every four terms.Step 4: Since the sum repeats every 4 terms and we are summing from $n = 1$ to $n = 24$, we have 6 full cycles of the 4-term pattern. The sum of one full cycle is:$$(i - 1) + (-1 - i) + (1 - i) + (1 + i) = 0.$$Thus, each cycle contributes 0 to the sum.Step 5: Since there are 6 full cycles, the total sum is: $$6 \times 0 = 0.$$Thus, the value of the sum is $0$.Therefore, the correct answer is option (D).Quick Tip: When dealing with powers of $i$, recognize the periodicity every four terms. This can simplify evaluating sums or expressions involving powers of $i$.