In (1−2x)26: T0=1, T1=−52, T2=1300, T3=−20800.
Coefficient of x: b−52c=−56 ...(1)
Coefficient of x2: a−52b+1300c=0 ...(2)
Coefficient of x3: −52a+1300b−20800c=0 ...(3)
From (1): b=52c−56. Substituting in (2): a=1404c−2912.
Substituting in (3): −26208c+78624=0⇒c=3.
Thus b=100, a=1300.
a+b+c=1300+100+3=1403.