Formula used:

where,

Here, a = 1 and b = 2

Therefore,

Let,

Here, f(x) = x² – 1 and a = 1

Now, by putting x = 1 in f(x) we get,

f(1) = 1² – 1 = 1 – 1 = 0

f(1 + h)

= (1 + h)² – 1

= h² + 1² + 2(h)(1) – 1

= h² + 2(h)

Similarly, f(1 + 2h)

= (1 + 2h)² – 1

= (2h)² + 1² + 2(2h)(1) – 1

= (2h)² + 2(2h)

{∵ (x + y)² = x² + y² + 2xy}

Now take h² and 2h common in remaining series

Put,

Since,