Question

if(x
)=x^100+x^99+....+x+1,thenf'(1) is equal to​

Answer 1

Answer:

f(x)=x100 + x99 + ………….x + 1

f'(x) = 100x99 + 99x98 +…… + 1

f'(1) = 100(1)99 + 99(1)98 + …… + 1

[Using Arithmetic Progression, where d = -1, a = 100 & n = 100]

= 50 (200 – 99)

= 50 (101)

= 5050