Question

Kindly answer quickly​

Answer 1

0 is the answer i think hopes it will helps you ok mark me as a brainlist and follow me

Answer 2

Answer:

(3)  0

Step-by-step explanation:

This is based on the Taylor series:

$\displaystyle f(x)=f(1) + \frac{f'(1)}{1!}(x-1) + \frac{f''(1)}{2!}(x-1)^2+\cdots\\\\\Rightarrow f(x+1) = f(1) + \frac{f'(1)}{1!}x+\frac{f''(1)}{2!}x^2+\cdots$

[ Notice that the expansion only goes up to the n-th derivative because here f(x) is a polynomial of degree n, so all higher derivatives in the Taylor series are equal to 0. ]

The value that we want corresponds to the series above with x = -1, to get the alternating signs.  So the value is

 f(-1+1) = f(0) = 0ⁿ = 0