Chapter 4 Class 11 Mathematical Induction
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 21 Important Deleted for CBSE Board 2024 Exams
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Chapter 4 Class 11 Mathematical Induction
Last updated at April 16, 2024 by Teachoo
Question 1 Prove the following by using the principle of mathematical induction for all n ∈ N: 1 + 3 + 32+……+ 3n – 1 = ((3𝑛 − 1))/2 Let P(n) : 1 + 3 + 32+……+ 3n – 1 = ((3𝑛 − 1))/2 Proving for n = 1 For n = 1, L.H.S = 1 R.H.S = ((3^1 − 1))/2 = ((3 − 1))/2 = ((2))/2 = 1 Since, L.H.S. = R.H.S ∴ P(n) is true for n = 1 Proving P(k + 1) is true if P(k) is true Assume that P(k) is true, P(k): 1 + 3 + 32 +…..+ 3k – 1 = ((3𝑘 − 1))/2 We will prove that P(k + 1) is true. P(k + 1): 1 + 3 + 32 +…..+ 3(k + 1) – 1 = ((3^(𝑘+1) − 1))/2 P(k + 1): 1 + 3 + 32 +…..3(k – 1) + 3(k) = ((3^(𝑘+1) − 1))/2 We have to prove P(k + 1) is true Solving LHS 1 + 3 + 32 +…..+ 3k – 1 + 3k From (1): 1 + 3 + 32 +…..+ 3k – 1 = ((𝟑𝒌 − 𝟏))/𝟐 = ((𝟑𝒌 − 𝟏))/𝟐 + 3k = ((3𝑘 − 1) + 2 × 3^𝑘)/2 = (𝟑𝒌 + 𝟐 × 𝟑^𝒌 − 𝟏)/𝟐 = ( 3(3^𝑘 )− 1)/2 = (𝟑^(𝒌 + 𝟏) − 𝟏)/𝟐 = RHS ∴ P(k + 1) is true when P(k) is true Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number