Question

plz rply with whole steps....i will definitely mark as brainliest​

Answer 1

Answer:

(A) 0.

Step-by-step explanation:

The square of a real number is zero only when the real number itself is zero.

So, for:

$(x - 1)^2 + (x - 2)^2 + (x - 3)^2 + ... + (x - n)^2$ to be zero, all of $x - 1$, $x - 2$, ..., $x - n$ must be zero.

So, x = 1, x = 2, ..., x = n, BUT at these values, only a single bracket becomes zero, and for NO value are all the terms zero together (since, fundamentally, x - 1 = 0 and x - 2 = 0 cannot be solved simultaneously).