Formula used:

where,

Here, a = 1 and b = 2

Therefore,

Let,

Here, f(x) = x² and a = 1

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

f(1) = 1² = 1

f(1 + h)

= (1 + h)²

= h² + 1² + 2(h)(1)

= h² + 1 + 2(h)

Similarly, f(1 + 2h)

= (1 + 2h)²

= (2h)² + 1² + 2(2h)(1)

= (2h)² + 1 + 2(2h)

{∵ (x + y)² = x² + y² + 2xy}

In this series, 1 is getting added n times

Now take h² and 2h common in remaining series

Put,

Since,