Generalized term be (2r - 1)²=4r² + 1 - 4r

1^st term = 4(1)² + 1 - 4(1)

2^nd term =4(2)² + 1 - 4(2)

And so on

nth term= 4n² + 1 - 4n

Summation=1^st term + 2^nd term + …….. + nth term

= 4(1)² + 1 - 4(1) + 4(2)² + 1 - 4(2) ………4n² + 1 - 4n ……(1)

We know ,

Thus

From (1) we have

Summation =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

Thus

(2)

We know

Thus substituting above values in (2)