Question 50 of 51advanced🔍 AnalyzeLong Answer5 marks

Suppose P1=1P_1 = 1, P2=2P_2 = 2 and for n>2n > 2, Pn=P1+P2+⋯+Pn−1+1P_n = P_1 + P_2 + \cdots + P_{n-1} + 1. Find the values of P1P_1, P2P_2, ..., P8P_8. Can you find a simpler recursive formula for PnP_n? Can you give an explicit formula?

Correct Answer

(Explicit formula) P_n = 2n−12^{n-1}

(Simpler recursive formula) P_n = 2P_{n-1} for n >= 2

(Values of P_1 to P_8) 1, 2, 4, 8, 16, 32, 64, 128

Exercise: END-OF-CHAPTER EXERCISES | Q: 14 | (Page 23)
For More Understanding

Explanation

The question requires calculating sequence terms using a given sum-based recursive formula, identifying the resulting geometric progression pattern, and expressing it in simpler recursive and explicit forms.

The context explains how sequences with a constant multiplier (common ratio) are geometric progressions, showing examples like 3,6,12,…3, 6, 12, \ldots with recursive formula tn=2tn−1t_n = 2t_{n-1} and explicit formula tn=3×2n−1t_n = 3 \times 2^{n-1}.

Applying this logic to the calculated PnP_n values reveals a common ratio of 22.

Solution Steps

  1. Step 1: Calculate P3=P1+P2+1=1+2+1=4P_3 = P_1 + P_2 + 1 = 1 + 2 + 1 = 4.

  2. Step 2: Calculate P4=P1+P2+P3+1=1+2+4+1=8P_4 = P_1 + P_2 + P_3 + 1 = 1 + 2 + 4 + 1 = 8.

  3. Step 3: Calculate P5=1+2+4+8+1=16P_5 = 1 + 2 + 4 + 8 + 1 = 16, P6=32P_6 = 32, P7=64P_7 = 64, P8=128P_8 = 128.

  4. Step 4: Identify the pattern: each term is twice the previous term (common ratio =2= 2), so simpler recursive formula is Pn=2Pn−1P_n = 2P_{n-1}.

  5. Step 5: Identify explicit formula for GP with a=1a = 1, r=2r = 2: Pn=2n−1P_n = 2^{n-1}.

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