if value of 1+∑_n=1⁵¹ (4n²+2n-1)((2n-1)!)=(53+a)! then a is equal to

Hi,

U need to rewrite this equation extensively.
Split the quadratic into 2n.2n and 2n-1
So when u multiply the factorial ull get 2n.2n.(2n-1)! + (2n-1)[(2n-1)!]
I can write like (2n+1-1).(2n!) + (2n-1+1-1)[(2n-1)!]
So it will become (2n+1)! – 2n! + 2n! – (2n-1)!
So finally we get (2n+1)!-(2n-1)!

Now start putting the values of summation, all the terms will cancel out except two which are 103!-1!

Now remember there is a 1+ in the begining of the question so we get only 103! which gives a=50