Proof the statement:2+4+6+....+2n=n(n+1)?

Question

it is by mathematical induction.

Answer 1

There are easier ways to show the result than by induction, but induction should work fine anyway.

If n = 1, then LHS = 2, RHS = 1(1 + 1) = 2 = LHS.

Assume equation holds for n. Need to prove for n+1:

LHS = 2 + ... + 2n + 2(n+1)
= n(n+1) + 2(n+1) by induction hypothesis
= (n +2)(n+1) by factorization
= ((n+1) + 1)(n+1)
= (n+1)((n+1) + 1)

QED.

Answer 2

It's just the triangle number series, twice over. You could prove it graphically, by drawing rows of little squares making a triangle and then rearranging them to produce a rectangle ..... but you asked for induction, so here goes.

The nth value, according to the RHS, is
n * (n + 1) == n**2 + n

The n+1th value, according to the RHS, is
(n + 1) * (n + 2) == n * (n + 1) + 2 * (n + 1)

So we have added 2* (n + 1) on the RHS, which is the same as what we have added on the LHS. This does not complete the proof: we only know that *if* the relationship holds for n, then it must also hold for n+1. To be sure that it does actually hold, we must select some arbitrary value for n and test the relationship.

When n = 3: on the LHS, 2 + 4 + 6 = 12
On the RHS, 3 * (3 + 1) = 3 * 4 = 12
= LHS

Therefore, 2 + 4 + 6 + ... + 2 * n = n * (n + 1). QED.

Answer 3

p(2): 2=1(1+1)True
p(k):2+4+6+....+2k=k(k+1)suppose True
p(k+1): 2+4+6+..+2k+2(k+1)=(k+1)(k+2)
in p(k+1)replace 2+4+6+...+2k with k(k+1) from p(k)
so k(k+1)+2(k+1)=(k+1)(k+2) divide all by (k+1)
k+2=k+2 True so p(n) True