Define mathematical induction :
Mathematical Induction is a method or technique of proving mathematical results or theorems
The process of induction involves the following steps.
Question 1 :
By the principle of mathematical induction, prove that, for n ≥ 1
1^{2} + 3^{2} + 5^{2} + · · · + (2n − 1)^{2} = n(2n − 1)(2n + 1)/3
Solution :
Let p(n) = 1^{2} + 3^{2} + 5^{2} + · · · + (2n − 1)^{2} = n(2n − 1)(2n+1)/3
Step 1 :
put n = 1
p(1) = 1^{2} + 3^{2} + 5^{2} + · · · + (2(1) − 1)^{2} = 1(2(1) − 1)(2(1)+1)/3
1 = 1
Hence p(1) is true.
Step 2 :
Let us assume that the statement is true for n = k
p(k) = 1^{2}+3^{2}+5^{2}+ · · · + (2k − 1)^{2} = k(2k − 1)(2k+1)/3 ---(1)
We need to show that P(k + 1) is true. Consider,
Step 3 :
Let us assume that the statement is true for n = k + 1
p(k+1)
1^{2 }+ 3^{2 }· · · + (2(k+1) − 1)^{2} = [(k+1)(2(k+1) − 1)(2(k+1)+1)]/3
1^{2 }+ 3^{2 }· · · + (2k+1)^{2} = [(k+1)(2k + 1) (2k + 3)]/3
1^{2 }+ 3^{2 }· · · + (2k-1)^{2 }+ (2k+1)^{2} = [(k+1)(2k + 1) (2k + 3)]/3
By applying (1) in this step, we get
Hence, by the principle of mathematical induction,n ≥ 1
1^{2} + 3^{2} + 5^{2} + · · · + (2n − 1)^{2} = n(2n − 1)(2n + 1)/3
Question 2 :
Prove that the sum of the first n non-zero even numbers is n^{2} + n.
Solution :
Let p(n) be the statement "n^{2} + n" is even.
Step 1 :
p(n) = n^{2} + n
put n = 1
p(1) = 1^{2} + 1 = 2, which is even
Hence p(1) is true.
Ste 2 :
Let p(m) is true. Then
p(m) is true ==> m^{2} + m is even ==> m^{2} + m ==> 2λ for some λ ∊ N
Now, we shall show that p(m + 1) is true. For this we have to show that (m+1)^{2} + (m + 1) is an even natural number.
Now,
(m+1)^{2} + (m + 1) = m^{2} + 2 m + 1 + m + 1
= m^{2} + 2 m + m + 2
= m^{2} + m + 2m + 2
= m^{2} + m + 2(m + 1)
= 2λ + 2 (m + 1)
= 2(λ + m + 1)
Hence p(m + 1) is true.
Hence the sum of the first n non-zero even numbers is n^{2} + n.
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