Question

Tính tổng sau: A=1+2^2+2^4+2^6+…+2^98+2^100

Answer 1

CÓ `A = 1 +2^2 + 2^4 +... + 2^100`

`=> 2^2A = 2^2 + 2^4 + ... + 2^102`

`=> 4A - A = (2^2 + 2^4 + ... + 2^102) - (1 + 2^2 + 2^4 + ... + 2^100)`

`=> 3A = 2^102 - 1`

`=>  A = (2^102 - 1)/3 `

Vậy `A = (2^102 - 1)/3`

Answer 2

\(A=1+2^2+2^4+2^6+...+2^{98}+2^{100}\)

\(4A=2^2+2^4+2^6+2^8+...+2^{100}+2^{102}\)

\(4A-A=2^2+2^4+2^6+2^8+...+2^{100}+2^{102}-1-2^2-2^4-2^6-...-2^{100}\)

\(3A=2^{102}-1\)

\(A=\dfrac{2^{102}-1}{3}\)

Vậy...

Answer 3

\(A=1+2^2+2^4+...+2^{100}\)

=>\(4A=2^2+2^4+...+2^{102}\)

=>\(4A-A=2^2+2^4+...+2^{102}-1-2^2-...-2^{100}\)

=>\(3A=2^{102}-1\)

=>\(A=\dfrac{2^{102}-1}{3}\)