Question on Proving a Sequence by Mathematical Induction

Topic : Mathematical Induction

Statement : Prove 2 + 4 +6 + 8 + . . . . . + 2n = n (n+1) by Mathematical Induction

Proof :

Let the statement be P(n),
P(n) : 2 + 4 + 6 + 8 + …….+2n = n(n+1)
P(1) : 2 = 1(1+1) = 1(2) = 2
2 = 2

So P(1) is true
Now let’s assume that P(k) is true

P(k) : 2 + 4 + 6 + 8 + …….+2k = k(k+1)-----------(1)
Now we have to prove P(k+1) is also true.

P(k+1) : 2 + 4 + 6 + 8 + …….+2(k+1) = (k+1)(k+2)

2 + 4 + 6 + 8 + …….2k+2(k+1) = (k+1)(k+2)-------(2)
we know 2 + 4 + 6 + 8 + …….2k = k(k+1) from (1)
So (2) will be
k (k+1) + 2(k+1) = (k+1)(k+2)

Taking (k+1) common
(k+1) + (k+2) = (k+1)(k+2)
Which is true.
So P(k+1) is also true.

Hence proved P(n) : 2 + 4 + 6 + 8 + …….+2n = n(n+1)