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2025 PEA Q21

f:N+→N+f : \mathbb{N}_+ \to \mathbb{N}_+, f(x+y)=f(x)f(y)f(x+y) = f(x) f(y), f(1)=2f(1) = 2. Find 2+∑n=12025f(n)2 + \sum_{n=1}^{2025} f(n).

Reveal answer and solution

Answer

B

Solution

  1. 1

    By induction, f(n)=f(1)n=2nf(n) = f(1)^n = 2^n. Therefore,

  2. 2
    ∑n=12025f(n)=∑n=120252n=22026−2. \sum_{n=1}^{2025} f(n) = \sum_{n=1}^{2025} 2^n = 2^{2026} - 2.
  3. 3

    Adding 22,

  4. 4
    2+∑n=12025f(n)=22026. 2 + \sum_{n=1}^{2025} f(n) = 2^{2026}.
  5. 5

    Hence the answer is 220262^{2026}.

Answer structure / marking notes

The geometric sum ∑n=1N2n=2N+1−2\sum_{n=1}^{N} 2^n = 2^{N+1} - 2 is the key identity. The constant 22 in the question cancels the −2-2 from the GP, yielding the clean answer 220262^{2026}.

\textit{Correction note: option (B) 220262^{2026} matches; the listed answer in the key is (B). Some answer keys mis-print this as (D); after explicit calculation, (B) is correct.}

Option B: 22026 \boxed{\text{Option B: } 2^{2026}}

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.