Formula used:

where,

Here, a = 1 and b = 4

Therefore,

Let,

Here, f(x) = 3x² + 2x and a = 1

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

f(1) = 3(1)² + 2(1) = 3 + 2 = 5

f(1 + h)

= 3(1 + h)² + 2(1 + h)

= 3{h² + 1² + 2(h)(1)} + 2 + 2h

= 3h² + 3 + 6h + 2 + 2h

= 3h² + 8h + 5

Similarly, f(1 + 2h)

= 3(1 + 2h)² + 2(1 + 2h)

= 3{(2h)² + 1² + 2(2h)(1)} + 2 + 4h

= 3(2h)² + 3 + 6(2h) + 2 + 2(2h)

= 3(2h)² + 8(2h) + 5

{∵ (x + y)² = x² + y² + 2xy}

Since 5 is repeating n times in series

Now take 3h² and 8h common in remaining series

Put,

Since,