Question

Tính hợp lý nếu có thể:

* A = 3 + 3² + 3³ + ... + 3¹⁰⁰

* B = 1 - 2 + 2² - 2³ + ... - 2⁹⁹ + 2¹⁰⁰

* C = x² - yz

* D = x . y² - z

* E = (x² - y²) . z

với x = -7; y = -3; z = 5

Answer 1

\(C=x^2-yz\)

\(=\left(-7\right)^2-\left(-3\right).5\)

\(=49+15=64\)

\(D=xy^2-z\)

\(=\left(-7\right).\left(-3\right)^2-5\)

\(=\left(-7\right).9-5\)

\(=-63-5=-68\)

\(E=\left(x^2-y^2\right).z\)

\(=\left[\left(-7\right)^2-\left(-3\right)^2\right].5\)

\(=\left(49-9\right).5\)

\(=40.5=200\)

Answer 2

\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3\left(3+3^2+3^3+...+3^{100}\right)\)

\(=3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-3\)

\(A=\dfrac{3^{101}-3}{2}\)

Answer 3

\(B=1-2+2^2-2^3+...-2^{99}+2^{100}\\ 2B=2-2^2+2^3-2^4+...-2^{100}+2^{101}\\ 2B+B=\left(2-2^2+2^3-2^4+...-2^{100}+2^{101}\right)+\left(1-2+2^2-2^3+...-2^{99}+2^{100}\right)\\ 3B=2^{101}+1\\ B=\dfrac{2^{101}+1}{3}\)