Great question. Great answers already by many.
Combinatorial argument for 1.
This answer is plagiarized from math.stackexchange.com so I deserve no credit at all. Just here to explain.
n^2 is the no. of integers less than 100.
2C(n, 2) : is the no. of numbers less than 100 having distinct digits.
C(n,1): is the no. of numbers less than hundred having both digits same.
Therefore,. n^2 = 2C(n, 2) + C(n,1)
n^2 = 2C(n,2) + n.
Now coming to the real part.
C(n, 2) + C( n+1, 2 ) = n^2
C(n,2) + C( n+1, 2 ) = 2C(n, 2) + n.
C( n+1, 2) = C(n, 2) + n. (we will prove that this is true.)
C( n+1, 2) = no. of edges in a n+1 complete graph.
C(n, 2) + n = no. of edges between n nodes + put another node and join it with all n nodes.
Therefore we have proved that,
C( n+1, 2 )= C(n, 2) + n.
C(n,2)+ C(n+1, 2) = n^2 is true.