Formula used:

where,

Here, a = 1 and b = 3

Therefore,

Let,

Here, f(x) = 2x² + 5x and a = 1

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

f(1) = 2(1)² + 5(1) = 2 + 5 = 7

f(1 + h)

= 2(1 + h)² + 5(1 + h)

= 2{h² + 1² + 2(h)(1)} + 5 + 5h

= 2h² + 4h + 2 + 5 + 5h

= 2h² + 9h + 7

Similarly, f(1 + 2h)

= 2(1 + 2h)² + 5(1 + 2h)

= 2{(2h)² + 1² + 2(2h)(1)} + 5 + 10h

= 2(2h)² + 2 + 8h + 5 + 10h

= 2(2h)² + 18h + 7

= 2(2h)² + 9(2h) + 7

{∵ (x + y)² = x² + y² + 2xy}

In this series, 7 is getting added n times

Now take 2h² and 9h common in remaining series

Put,

Since,