Problem 1 :
Use induction to prove that n^{3} − 7n + 3, is divisible by 3, for all natural numbers n.
Solution :
Let P(n) = n^{3} – 7n + 3 is divisible by 3, for all natural numbers n.
Step 1 :
Now P(l): (l)^{3} – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Step 2 :
Let us assume that P(n) is true for some natural number n = k.
P(k) = K^{3} – 7k + 3 is divisible by 3
or K^{3} – 7k + 3 = 3m, m∈ N (i)
Step 3 :
Now, we have to prove that P(k + 1) is true.
P(k+ 1 )(k + l)^{3} – 7(k + 1) + 3
= k^{3} + 1 + 3k(k + 1) – 7k— 7 + 3
= k^{3} -7k + 3 + 3k(k + l) - 6
= 3m + 3[k(k+l)-2] [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.
Problem 2 :
Use induction to prove that 10^{n} + 3 × 4^{n+2} + 5, is divisible by 9, for all natural numbers n.
Solution :
Step 1 :
n = 1 we have
P(1) ; 10 + 3 ⋅ 64 + 5 = 207 = 9 ⋅ 23
Which is divisible by 9 .
P(1) is true .
Step 2 :
For n =k assume that P(k) is true .
Then P(k) : 10^{k} + 3.4 ^{k+2} + 5 is divisible by 9.
10^{k} + 3.4^{k+2} + 5 = 9m
10^{k} = 9m - 3.4^{k+2} - 5 ----------(1)
Step 3 :
We have to to prove that P(k+1) is divisible by 9 for
n = k + 1.
P(k + 1) : 10^{k+1} + 3.4^{k+1+2} + 5
= 10 x 10^{k} + 3.4^{k+2} .4+ 5
= 10 ( 9m - 3.4^{k+2} - 5 ) + 3.4^{k+2} .4+ 5
= 90m - 30 4^{k+2} - 50 + 12.4^{k+2} + 5
= 90m - 18 4^{k+2} - 45
= 9(10m - 2.4^{k+2} - 5 )
which is divisible by 9 .
P (k +1 ) is true .
Hence by the Principle of mathematical induction P(n) is true for all n∈N.
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