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Example 7 - Prove that 12 + 22 + ... + n2 > n3/3 - Induction - Examples

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Example 7 - Chapter 4 Class 11 Mathematical Induction - Part 2

Example 7 - Chapter 4 Class 11 Mathematical Induction - Part 3
Example 7 - Chapter 4 Class 11 Mathematical Induction - Part 4
Example 7 - Chapter 4 Class 11 Mathematical Induction - Part 5

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Example 7 Prove that 12 + 22 + ... + n2 > n3/3 , n ∈ N Introduction Since 10 > 5 then 10 > 4 + 1 then 10 > 4 We will use the theory in our question Example 7 Prove that 12 + 22 + ... + n2 > n3/3 , n ∈ N Let P(n) : 12 + 22 + ... + n2 > n3/3 , n ∈ N For n = 1 L.H.S = 12 = 1 R.H.S = 13/3 = 1/3 Since 1 > 1/3 L.H.S > R.H.S ∴ P(n) is true for n = 1 Assume P(k) is true P(k) : 12 + 22 + ... + k2 > k3/3 We will prove that P(k + 1) is true. L.H.S = 12 + 22 + 32 + ... + (k + 1)2 R.H.S = (k + 1)^3/3 L.H.S = 12 + 22 + 32 + ... + (k + 1)2 = 12 + 22 + 32 + ... + k2+ (k + 1)2 = (12 + 22 + 32 + ... + k2 )+ (k + 1)2 Using (1): 12 + 22 + ... + k2 > 𝑘3/3 > k3/3 + (k+1)2 > (𝑘^3+3(𝑘+1)^2)/3 > 1/3 [k3 + 3(k2 + 2k + 1)] > 1/3 [k3 + 3k2 + 6k + 3] > 1/3 [ ( k3 + 1 + 3k2 + 3k)+ (3k+2)] > 1/3 (k3 + 1 + 3k2 + 3k ) + 1/3 (3k+2) As 1/3 (3k+2) is a positive quantity > 1/3 { k3 + 1 + 3k2 + 3k } R.H.S = (k + 1)^3/3 Using (a+b)3 = a3 + b3 +3a2b+3ab2 = 1/3 (k3 + 13 + 3k + 3k2 ) = 1/3 (k3 + 1 + 3k + 3k2 ) L.H.S > R.H.S ∴ P(k + 1) is true whenever P(k) is true. ∴By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.