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For n = 1
LHS of P(n) = 1
since, LHS = RHS
⟹ p(n)is true for n = 1
Let P(n) be true for n = k, so
Now
⟹p(n) is true for n = k + 1
⟹p(n) is true for all n ϵ N
so, by the principle of mathmematical induction
1 + 1
⟹ p(n) is true for n = 1
Let p(n) is true for n = k, so
We have to show that p(n)is true for n = k + 1
⟹ p(n) is true for n = k + 1
⟹ p(n) is true for all n ϵ N by PMI
1 = 1
⟹p(n) is true for n = 1
Let p(n) is true for n = k
we have to show P(n) is true for n = k + 1
Now,
we have to show that
⟹ p(n) is true for n =k + 1
Let p(n) : 1 + 3 + 5 + ...+ (2n - 1) = n2
p(1) : 1 = 12
p(k) : 1 + 3 + 5 +...+ (2k-1) = k2 ----(1)
1 + 3 + 5 +....+(2k - 1) +2(k + 1) -1 = (k + 1)2
{1+3+5+..+(2k - 1)} + (2k + 1)
= k2 + (2k + 1) [Using equation (1)]
= k2 + 2k + 1
= (k + 1)2
Put n = 1
We have to show that,
P(n) is true for n = k + 1
P(n) is true for all n ϵ N by PMI
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Chapter 12: Mathematical Induction –...