Question

tính A=2⁰+2³+2⁵+....+2⁹⁹

Answer 1

\(A=2^0+2^3+2^5+...+2^{99}.\)

\(4A=4.\left(2^0+2^3+2^5+...+2^{99}\right)\)

\(4A=2^2+2^5+2^7+....+2^{99}+2^{101}\)

\(4A-A=2^{101}-2^2\)

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

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

Answer 2

Số các số hạng là : 2^99 - 2^0 = 2^98

Tổng là : (2^99 + 2^0) x 2^98 = 2^198

Answer 3

A=2⁰+2³+2⁵+....+2⁹⁹

2A = 2³ + 2⁵ + ... + 2¹⁰⁰

2A - A = ( 2⁰ + 2³ + 2⁵ + ... + 2¹⁰⁰ ) - ( 2⁰ + 2³ + 2⁵ + .... + 2⁹⁹ )

1A = 2¹⁰⁰ - 2

\(A=\frac{2^{100}-2}{1}\)

Answer 4

A = 2⁰ + 2³ + 2⁵ + ... + 2⁹⁹

2³ . A = 2³ + 2⁵ + 2⁷ + ... + 2¹⁰¹

2³A - A = ( 2³ + 2⁵ + 2⁷ + ... + 2¹⁰¹ ) - ( 2⁰ + 2³ + 2⁵ + ... + 2⁹⁹ )

7A = 2¹⁰¹ - 2⁰

A = ( 2¹⁰¹ - 2⁰ ) : 7