by induction, for every positive integer n, 1 + 3 + 5 +. + (2n - 1) = n2. Q.
Principle of Mathematical Induction: If P is a set of integers such that (i) a is in P, (ii) for all k ≥ a, if the integer k is in P, then the integer k + 1 is also in P, then P = {x ∈ Z | x ≥ a} that is, P is the set of all integers greater than or equal to a.
Answers & Comments
We need to show that 1 + 2 + 3 + ... + n = n(n + 1)/2 using mathematical induction.
According to the ways of mathematical induction, we first need to check for the initial cases (Strong+rule of mathematical induction).
For n = 1, we have 1(1 + 1)/2 = 1.
For n = 2, we have 2(2 + 1)/2 = 3 (which is 1 + 2).
For n = 3, we have 3(3 + 1)/2 = 6.
Now, suppose the results holds for n = k,.i.e,
1 + 2 + 3 + ... + k = k(k + 1)/2.
The next thing is to show that the result holds for k + 1.i.e,
1 + 2 + 3 + ... + (k + 1) = (k + 1)[(k + 1) + 1)]/2 = (k + 1)(k + 2)/2.
To show this, let's use our information for n = k;
1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1)
= (k + 1)[(k/2 + 1)] = (k + 1)(k + 2)/2.
We are done.
by induction, for every positive integer n, 1 + 3 + 5 +. + (2n - 1) = n2. Q.
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