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Problem 1. Prove that for all positive integers n, we have 1 + 3 + ....

Problem 1. Prove that for all positive integers n, we have 1 + 3 + . . . + (2n − 1) = n ^2 .

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Answer #1

P(n): 1 + 3 + 5 + ...+(2n-1)=n 2
For n=1
P(1) = 2n-1<
= 2(1) -1
= 2 -1 =1
n 2 = (1) 2
∴ P(1) is true. ----(equation 1)
Assume that P(k) is true.
So P(k): 1 + 3 + 5 +....(2k -1) = k 2 is true. ----(equation 2)
Now we have to prove that P(k+1) is true.
P(k+1) : 1 + 3 + 5 +....+(2k-1) + [2(k+1)-1]
= k 2 + (2k + 1)
= (k+1) 2
∴ P(k+1) is true whenever p(k) is true.
Hence by equation (1), (2) and by principle of mathematical induction P(n) is true for all natural number 'n'.

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