Fill in the missing pieces in the following proof that
1 + 3 + 5 + … + (2n –1) = n
for all integers n > 1.
Proof: Let the property P (n) be the equation
1 + 3 + 5 + +(2n –1) = n2. ← P(n)
Show that P(1) is true: To establish P (1), we must show that when 1 is substituted in place of n, the left-hand side equals the right-hand side. But when n = 1, the left-hand side is the sum of all the odd integers from 1 to 2•1 – 1, which is the sum of the odd integers from 1 to 1, which is just 1. The right-hand side is (a) which also equals 1. So P(1 ) is true.
Show that for all integers k > 1, if P (k) is true then P (k +1) is true: Let k be any integer with k ≥ 1.
[Suppose P (k) is true. That is:]
Suppose 1 + 3 + 5 + … + (2k – 1) =
[This is the inductive hypothesis.]
[We must show that P (k + 1) is true. That is:]
We must show that
But the left-hand side of P (k + 1) is
which is the right-hand side of P (k + 1) [as was to be shown.]
[Since we have proved the basis step and the inductive step, we conclude that the given statement is true.]Publisher information found
The previous proof was annotated to help make its logical flow more obvious. In standard mathematical writing, such annotation is omitted.