QC 1 on Mathematical Induction in 'Walk through Combinatorics'

Question

Use induction to prove that if a_n = 1+2+···+n, then a_n = n(n+1)/2, for all positive integers n.

Answer 1

The Initial Step. a₁ = 1 and a₁ = 1(2)/2 = 1. Thus, the equation is true for n=1.

The Inductive Step. Let us assume the equation is true for n. Then,
1+2+...+n = n(n+1)/2 ----1

Now, we have to prove the equation for n+1. i.e.
1+2+...+n+n+1 = (n+1)(n+2)/2 ----2

eqn 2-1, we get
n+1 = (n+1)(n+2)/2 - n(n+1)/2 ----3
This is a true statement, and we know by inductive step that eqn 1 is a true statement. Adding both these statements, we get, 1+2+..+n+n+1 = n(n+1)/2 + (n+1)(n+2)/2 - n(n+1)/2 = (n+1)(n+2)/2
Hence Proved.