Question

Cho f(x)= x¹⁰⁰ + x⁹⁹ + x⁹⁸ + ..... + x² + x+ 1. Tính f(1), f(-1), f(2), f(-2)

Cho g(x)= x + x³ + x⁵ + ..... + x¹⁰¹. Tính g(1), g(-1), g(3)

Answer 1

*) f(1) = 1^100 + 1^99 + ...+ 1 + 1

= 1+ 1 + 1 + ...+ 1 + 1 (101 số 1)

= 101

tương tự:

*) f(-1) = -1 - 1 - 1 ... - 1 - 1 + 1 (100 chữ số 1)

= -100 + 1 = -99

*) đặt f(2) = 2^100 + 2^99 + ...+ 2^2 + 2 + 1 = A

=> 2A = 2^101 + 2^100 + ... + 2^3 + 2^2 + 2

=> 2A - A = 2^101 + 2^100 + ... + 2^3 + 2^2 + 2 - ( 2^100 + 2^99 + ...+ 2^2 + 2 + 1)

<=> A = 2^101 - 1

=> f(2) = 2^101 - 1

tương tự:

*) đặt f(-2) = -2^100 - 2^99 ...- 2^2 - 2 - 1 = B

=> 2B = -2^101 - 2^100 ... - 2^3 - 2^2 - 2

=> 2B -B = -2^101 - 2^100 ... - 2^3 - 2^2 - 2 - ( -2^100 - 2^99 ...- 2^2 - 2 - 1)

<=> B = -2^101 + 1

=> f(-2) = -2^101 + 1

Answer 2

g(1) = 1 + 1^3 + 1^5 + ... + 1^101 (51 số 1)

= 51

g(-1) = -1 - 1^3 - 1^5.... - 1^101 (51 số 1)

= -51

đặt g(3) = 3 + 3^3 + 3^5 + ...+ 3^101 = A

=> 3^2 * A = 3^3 + 3^5 + ....+ 3^103

=> 9A - A = 3^3 + 3^5 + ....+ 3^103 - (3 + 3^3 + 3^5 + ...+ 3^101)

=> 8A = -3 + 3^103

=> A = \(\dfrac{3^{103}-3}{8}\)

=> g(3) = \(\dfrac{3^{103}-3}{8}\)