Question

find the remainer when 1+x+x^2+x^3+...+x^2014 divided by x-1 is

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm \longrightarrow f(x) = 1 + x + {x}^{2} + ... {x}^{2014}$

By remainder theorem - When f(x) is divided by (x - 1), the remainder is f(1).

Therefore:

$\rm = 1 + 1 + {1}^{2} + {1}^{3} + ... {1}^{2014}$

$\rm = 1 + 2014$

$\rm = 2015$

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

1. Remainder Theorem: When a polynomial f(x) is divided by (x - a), the remainder obtained is f(a).

2. Factor Theorem: Let f(x) be a polynomial of degree ≥ 2. Then, by factor theorem, (x - a) is a factor of f(x) iff f(a) = 0.

Answer 2

if x - 1 is the Factor.

so, x = 1

then,

⇒ 1 + 1 + 1² + 1³ + … + 1²⁰¹⁴

⇒ 1 + 2014

⇒ 2015 --> Required Answer