Formula used:

where,

Here, a = 0 and b = 2

Therefore,

Let,

Here, f(x) = x² + 2x + 1 and a = 0

Now, by putting x = 0 in f(x) we get,

f(0) = 0² + 2(0) + 1 = 0 + 0 + 1 = 1

f(h)

= (h)² + 2(h) + 1

= h² + 2h + 1

Similarly, f(2h)

= (2h)² + 2(2h) + 1

Since 1 is repeating n times in the series

Now take h² and 2h common in remaining series

Put,

Since,