Toán Lớp 6: chứng tỏ rằng 1+3+3^2+…+3^99 chia hết cho 40
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Toán Lớp 6: chứng tỏ rằng 1+3+3^2+…+3^99 chia hết cho 40
Comments ( 2 )
$\text{1 + 3 + 3² + 3³ + …. + $3^{99}$ }$
$\text{= ( 1 + 3 + 3² + 3³ ) + ( $3^{4}$ + $3^{5}$ + $3^{6}$ + $3^{7}$ ) + ….. + ( $3^{96}$ + $3^{97}$ + $3^{98}$ + $3^{99}$ )}$
$\text{= 1. ( 1 + 3 + 3² + 3³ ) + $3^{4}$. ( 1 + 3 + 3² + 3³ ) + …. + $3^{96}$. ( 1 + 3 + 3² + 3³ )}$
$\text{= ( 1 + $3^{4}$ + …. + $3^{96}$ ). ( 1 + 3 + 3² + 3³ )}$
$\text{= ( 1 + $3^{4}$ + …. + $3^{96}$ ). 40}$
$\text{Mà 40 $\vdots$ 40 nên ( 1 + $3^{4}$ + …. + $3^{96}$ ). 40 $\vdots$ 40.}$
$\text{Vậy 1 + 3 + 3² + 3³ + …. + $3^{99}$ $\vdots$ 40.}$
= 1(1+3+3^2+3^3) + 3^4(1+3+3^2+3^3)+ …. + 3^96(1+3+3^2+3^3)
= (1+3^4+…+3^96)(1+3+3^2+3^3)
= (1 + 3^4 + … + 3^96 ) x 40
=> Biểu thức trên chia hết cho 40
@bibihocbai