it is true for x = 1
5(1) - 3 = (1)[5(1) - 1]/2
5-3 = 4/2
2 = 2
If true for n, is it true for n+1?
n(5n-1)/2 = n(5n-1)/2
n(5n-1)/2 + [5(n+1) -3] = (n + 1)[5 (n+1) -1]/2
(5n2 - n)/2 + (5n + 5 -3) = [(5n2 + 10n +5) - (n+1)]/2
(5/2)n2 + (9/2)n + 2 = (5n2 + 9n +4)/2
for ease of calculation and recognition, just multiply both sides by 2
5n2 + 9n + 4 = 5n2 + 9n + 4
∴ if true for n, then true for n+1