Toán Tổng sau có chia hết cho 3 ko A= 2 + 2^2 + 2^3 +…+2^10 24/07/2021 By Melody Tổng sau có chia hết cho 3 ko A= 2 + 2^2 + 2^3 +…+2^10
A= 2 + 2^2 + 2^3 +…+2^10 A=(2+2^2)+(2^3+2^4)+……+(2^9+2^10) A=2.(1+2)+2^3.(1+3)+……..+2^9.(1+2) A=2.3+2^3.3+……+2^9.3 A=3.(2+2^3+…..+2^9) ma 3 chia het cho 3 ⇒A chia het cho 3 chuc bn hoc tot ! Trả lời
A = 2 + $2^{2}$ +$2^{3}$ + ……………+ $2^{10}$ = ( 2 + $2^{2}$ + ( $2^{3}$ + $2^{4}$ ) + …………………. + ( $2^{9}$ + $2^{10}$ ) = 2 . ( 1 + 2 ) + ……………………… + $2^{9}$ . ( 1 + 2 ) = 2 . 3 + …………………………………+ $2^{9}$ . 3 = 3 . ( 2 + $2^{9}$ ) chia hết cho 3 ⇒ A chia hết cho 3 . Trả lời
A= 2 + 2^2 + 2^3 +…+2^10
A=(2+2^2)+(2^3+2^4)+……+(2^9+2^10)
A=2.(1+2)+2^3.(1+3)+……..+2^9.(1+2)
A=2.3+2^3.3+……+2^9.3
A=3.(2+2^3+…..+2^9)
ma 3 chia het cho 3
⇒A chia het cho 3
chuc bn hoc tot !
A = 2 + $2^{2}$ +$2^{3}$ + ……………+ $2^{10}$
= ( 2 + $2^{2}$ + ( $2^{3}$ + $2^{4}$ ) + …………………. + ( $2^{9}$ + $2^{10}$ )
= 2 . ( 1 + 2 ) + ……………………… + $2^{9}$ . ( 1 + 2 )
= 2 . 3 + …………………………………+ $2^{9}$ . 3
= 3 . ( 2 + $2^{9}$ ) chia hết cho 3
⇒ A chia hết cho 3 .