Formula used:

where,

Here, a = 0 and b = 3

Therefore,

Let,

Here, f(x) = 2x² + 3x + 5 and a = 0

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

f(0) = 2(0)² + 3(0) + 5 = 0 + 0 + 5 = 5

f(h)

= 2(h)² + 3(h) + 5

= 2h² + 3h + 5

Similarly, f(2h)

= 2(2h)² + 3(2h) + 5

Since 5 is repeating n times in the series

Now take h² and 2h common in remaining series

Put,

Since,