Introductory Discrete Mathematics 2

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Introductory Discrete Mathematics Page 156 Question 4 b. n3+(n+1)3+(n+2)3Is divisible by 9 for all n≥1Proof The base case in this situation is n=0 which implies that n3+(n+1)3+(n+2)3=03+13+23=9 This is divisible by 9. Therefore, the basis holds true for n=0The inductive hypothesis becomes n=k≥0, k3+k+13+(k+2)3 is also divisible by 9 → k3+k+13+(k+2)3=9p…(i)By inductionn=k+1, equation (i) becomes (k+1)3+(k+2)2+(k+3)3=9p-k3+k+33…(ii) Simplifying (ii) we have =9p-k3+k+33=9p-k3+k3+9k2+9k+27=9p+9k2+k+3=9(p+k2+k+3)This implies that (k+1)3+(k+2)2+(k+3)3 is also divisible by 9 therefore, n3+(n+1)3+(n+2)3is divisible by 9 for all n≥1 d. 83 – 3n Is divisible by 5 for all n≥1Proof If n=1 then, 8n-3n= 81+31=5 this can be divided by 5. Therefore, the basis holds true for n=1The inductive hypothesis becomes n=k ≥1. It means that 8k-3k is divisible by 5 →8k-3k=5p…(i)By induction, taking n=k+1 we have, 8k+1-3k+1=88k-33k=38k-3k+5×8k…(ii)But 8k-3k=5p for some integer p. therefore, ii becomes =(3×5p)+(5×8k) Simplifying we get =5(3p+8k) Which is divisible by 5 implying that 8n – 3n is divisible by 5 for all n≥1 g. n3+5n Is divisible by 6 for all n≥1 Proof If n=1 then n3+5n=13+5×1=6, this is divisible by 6. Therefore, the basis holds true for n=1. By induction, we can assume that for some k, k3+5k can be divided by 6 Therefore, (k+1)3+5(k+1) should be divisible by 6. It also means that k3+5k=6p given some integer p. Expanding we have k+13+5k+1=k3+3k2+3k+1+5k+5=k3+3k2+8k+6=6p+1+3k2+3k (Since k3+5k=6p)…(i)In case k is an even number, then k=2x for some integer x. Substituting this value in equation (i) we

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