Formula used:

where,

Here, a = 2 and b = 3

Therefore,

Let,

Here, f(x) = 2x² + 1 and a = 2

Now, by putting x = 2 in f(x) we get,

f(2) = 2(2²) + 1 = 2(4) + 1 = 8 + 1 = 9

f(1 + h)

= 2(2 + h)² + 1

= 2{h² + 2² + 2(h)(2)} + 1

= 2(h)² + 8 + 2(4h) + 1

= 2(h)² + 9 + 8(h)

Similarly, f(2 + 2h)

= 2(2 + 2h)² + 1

= 2{2(2h)² + 2² + 2(2h)(2)} + 1

= 2(2h)² + 8 + 8(2h) + 1

= 2(2h)² + 9 + 8(2h)

{∵ (x + y)² = x² + y² + 2xy}

In this series, 9 is getting added n times

Now take 2h² and 4h common in remaining series

Put,

Since,