(2a+x)(1-3x)ⁿ                                                                                   =  4 -59x + bx² + …..

Using the Binomial expansion of:

(1+x)ⁿ =  1 + nx + n!/[(n-2)! 2! ] x² + ………    replacing x with (-3x)

(2a+x) { 1 + n(-3x) + n!/[(n-2)! 2! ] (-3x)²  + ……..}  = 4 -59x + bx² + …..

2a –(6an)x  +  n!/[(n-2)! 2! ](2a)(9x²) + (1) x – (3n) x^{2 + …….}    =  4 –59x + bx²  + …….

2a = 4 => a = 2

Comparing coefficient of x and substituting a =2,

6an – 1 = 59  => 12n = 60   => n = 5

Comparing coefficient of x² and substituting n = 5, a =2,

b = (5!/2!3!) (18×2) – 15 = 360 – 15 = 345

Therefore, a = 2, n = 5 and b = 345

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