Answer
k2 + 3k + 2,
Work Step by Step
Proof: For the given statement, the property P(n) is the
equation
← P(n)
2 + 4 + 6+· · ·+2n = $n^2$ + n.
Showing that P(1) is true:
To prove P(1),
- we must show that when 1 is substituted into the equation in place of n, the left-hand side equals the right-hand side.
-But when 1 is substituted for n, the left-hand side is the sum of all the even integers from 2 to 2· 1, which is just 2, and the right-hand side is 12 + 1,
which also equals 2.
Thus 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, and suppose P(k) is true.
That is, suppose 2 + 4 + 6+· · ·+2k = $k^2$ + k. ← P(k)
inductive hypothesis
We must show that P(k + 1) is true. That is, we must show that
2 + 4 + 6+· · ·+2(k + 1) = (k + 1)2 + (k + 1).
Because (k + 1)2 + (k + 1) = k2 + 2k + 1 + k + 1 = k2 + 3k + 2,
this is equivalent to showing that 2 + 4 + 6+· · ·+2(k + 1) = k2 + 3k + 2. ← P(k + 1)
But the left-hand side of P(k + 1) is
2 + 4 + 6+· · ·+2(k + 1) = 2 + 4 + 6+· · ·+2k + 2(k + 1)
by making the next-to-last term explicit = (k2 + k) + 2(k + 1) by substitution from the inductive hypothesis
by algebra,
= k2 + 3k + 2,
