The induction should be on the number of numbers, not the numbers themselves. Here's the proof you're looking for, for what it's worth:
The proof is by induction on the number of even numbers to be summed.
Base case: Let a
and b
be any two even numbers. Because a
and b
are even, we can write a = 2c
and b = 2d
for some integers c
and d
(not necessarily even). We then have a + b = 2c + 2d = 2(c + d)
. Since a + b = 2e
, where e = c + d
is some integer, then a + b
is even by definition.
Induction hypothesis: Assume that the sum of any collection of up to k >= 2
even numbers is even.
Induction step: We must show that the sum of any collection of k + 1
even numbers is also even. Let {n_1, n_2, ..., n_k, m}
be any collection of k + 1
even numbers. We must show that n_1 + n_2 + ... + n_k + m
is even. By the associative and commutative properties of addition, we can safely write this as (n_1 + n_2 + ... + n_k) + m
without changing the value of the expression. From the induction hypothesis, we know that p = n_1 + n_2 + ... + n_k
must be even, since it is the sum of a collection of up to k
even numbers. We now have p = 2x
, m = 2y
, and n_1 + n_2 + ... + n_k + m = p + m = 2x + 2y = 2(x + y) = 2z
for z = x + y
, and therefore we know that n_1 + n_2 + ... + n_k + m
must be even.
This concludes the proof by induction.