# Toán 7toán đại 7

#### Only Normal

##### Bá tước Halloween|Cựu TMod Toán
HV CLB Hóa học vui
Thành viên
Bài 2:
$A = 1+7+7^2+7^3+ ... + 7^{49} + 7^{50}$
$7A = 7+7^2+7^3+ ... + 7^{50} + 7^{51}$
$7A - A = ( 7+7^2+7^3+ ... + 7^{50} + 7^{51}) - (1+7+7^2+7^3+ ... + 7^{49} + 7^{50})$
$6A = 7^{51} -1$
$A = \dfrac{7^{51} -1}{6}$

phươnguyen080701

Thành viên

#### Only Normal

##### Bá tước Halloween|Cựu TMod Toán
HV CLB Hóa học vui
Thành viên
Bài 3 :
$B = (\dfrac{1}{2}) + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{98} + (\dfrac{1}{2})^{99}$
$\dfrac{1}{2}B = (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{99} + (\dfrac{1}{2})^{100}$
$\dfrac{1}{2}B - B = [ (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{99} + (\dfrac{1}{2})^{100}] - [(\dfrac{1}{2}) + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{98} + (\dfrac{1}{2})^{99}]$
$-0,5B = (\dfrac{1}{2})^{100} - \dfrac{1}{2}$
$B = \dfrac{(\dfrac{1}{2})^{100} - \dfrac{1}{2}}{-0,5}$
$B =( \dfrac{1}{2})^{100} - \dfrac{1}{2}:{\dfrac{-1}{2}}$
$B =- \dfrac{1}{2})^{100} - \dfrac{1}{2}.-2$
Vì $(- \dfrac{1}{2})^{100} - \dfrac{1}{2}.2 <1$
⇒ $B<1$

Last edited: