Cho S= 2^1 + 2^2 + 2^3 +…….+2^100 a) Tính S b) Chứng minh rằng S chia hết cho 3 06/07/2021 Bởi Amara Cho S= 2^1 + 2^2 + 2^3 +…….+2^100 a) Tính S b) Chứng minh rằng S chia hết cho 3
`a, S = 2 + 2^2 + 2^3 + … + 2^100` `→ 2S = 2^2 + 2^3 + 2^4 + … + 2^101` `→ 2S – S = (2^2 + 2^3 + 2^4 + … + 2^101) – (2 + 2^2 + 2^3 + … + 2^100)` `→ S = 2^101 – 2` `b, S = 2(2^100-1)` Ta có: `2^10 : 3` dư `1` `→2^100:3` dư `1` `→2^100 -1` chia hết cho `3` `→ S=2(2^100-1)` chia hết cho `3` Bình luận
$a.S=2+22+23+…+2100$$2S=22+23+24+…+2101$$2S−S=(22+23+24+…+2101)−(2+22+23+…+2100)$$S=2101−2$ $b.S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 …….. + 2^100$ $S = (2^1 + 2^2) +(2^3 + 2^4) + …………….. + (2^99 + 2^100)$ $S = 2^1.(1+2) + 2^3.(1+2) + 2^99.(1+2)$ $S= 2 . 3 + 2^3 . 3 + …. +2^99.3$ $S = 3 .(2 + 2^3 + ….+ 2^99) chia hết cho 3$ $ĐPCM$ $Cái này lớp 7 à nha$$Học$ $tốt!!!$ Bình luận
`a, S = 2 + 2^2 + 2^3 + … + 2^100`
`→ 2S = 2^2 + 2^3 + 2^4 + … + 2^101`
`→ 2S – S = (2^2 + 2^3 + 2^4 + … + 2^101) – (2 + 2^2 + 2^3 + … + 2^100)`
`→ S = 2^101 – 2`
`b, S = 2(2^100-1)`
Ta có: `2^10 : 3` dư `1`
`→2^100:3` dư `1`
`→2^100 -1` chia hết cho `3`
`→ S=2(2^100-1)` chia hết cho `3`
$a.S=2+22+23+…+2100$
$2S=22+23+24+…+2101$
$2S−S=(22+23+24+…+2101)−(2+22+23+…+2100)$
$S=2101−2$
$b.S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 …….. + 2^100$
$S = (2^1 + 2^2) +(2^3 + 2^4) + …………….. + (2^99 + 2^100)$
$S = 2^1.(1+2) + 2^3.(1+2) + 2^99.(1+2)$
$S= 2 . 3 + 2^3 . 3 + …. +2^99.3$
$S = 3 .(2 + 2^3 + ….+ 2^99) chia hết cho 3$
$ĐPCM$
$Cái này lớp 7 à nha$
$Học$ $tốt!!!$