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Last updated at March 29, 2023 by Teachoo

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