Giải thích các bước giải:

a/ $S=1+5+5^2+5^3+....+5^{100}$

⇒ $5S=5.(1+5+5^2+5^3+....+5^{100})$

⇒ $5S=5+5^2+5^3+....+5^{101}$

⇒ $5S-S=(5+5^2+5^3+....+5^{101})-(1+5+5^2+5^3+....+5^{100})$

⇒ $4S=5^{101}-1$

⇒ $S=\dfrac{5^{101}-1}{4}$

b/ $M=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{49}}$

⇒ $2M=2.(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{49}})$

⇒ $2M=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{48}}$

⇒ $2M-M=(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{48}})-(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{49}})$

⇒ $M=1-\dfrac{1}{2^{49}}$

⇒ $M=\dfrac{2^{49}-1}{2^{49}}$

c/ $N=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+......+\dfrac{1}{3^{100}}$

⇒ $3N=3.(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+......+\dfrac{1}{3^{100}})$

⇒ $3N=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{99}}$

⇒ $3N-N=(1+\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{99}})-(\dfrac{1}{3}+\dfrac{1}{3^2}+......+\dfrac{1}{3^{100}})$

⇒ $2N=1-\dfrac{1}{3^{100}}$

⇒ $2N=\dfrac{3^{100}-1}{3^{100}}$

⇒ $N=\dfrac{3^{100}-1}{2.3^{100}}$

d/ $D=7+7^2+7^3+....+7^{100}$

⇒ $7D=7.(7+7^2+7^3+....+7^{100})$

⇒ $7D=7^2+7^3+7^4+....+7^{101}$

⇒ $7D-D=(7^2+7^3+7^4+....+7^{101})-(7+7^2+7^3+....+7^{100})$

⇒ $6D=7^{101}-7$

⇒ $D=\dfrac{7^{101}-7}{6}$

Chúc bạn học tốt !!!

Đáp án:

 4.

a, Ta có : 

$S = 1 + 5 + 5^2 + 5^3 + .... + 5^{100} (1)$

$ => 5S = 5 + 5^2 + 5^3 + 5^4 + .... + 5^{101} (2)$

Lấy (2) trừ (1) ta đc

$4S = 5^{101} - 1$

$ => S = \dfrac{5^{101} - 1}{4}$ 

b, Ta có : 

M = $\dfrac{1}{2}$ + $\dfrac{1}{2^2}$ + $\dfrac{1}{2^3}$ + ..... + $\dfrac{1}{2^{49}}$ (1)

$ => 2M = 1 +$ $\dfrac{1}{2}$ + $\dfrac{1}{2^2}$  + ..... + $\dfrac{1}{2^{48}}$ (2)

Lấy (2) - (1) ta đc 

$M = 1 - \dfrac{1}{2^{49}} = \dfrac{2^{49} - 1}{2^{49}}$

c, Ta có : 

N = $\dfrac{1}{3}$ + $\dfrac{1}{3^2}$ + $\dfrac{1}{3^3}$ + ..... + $\dfrac{1}{3^{100}} (1)$

 $=> 3N =1 +$  $\dfrac{1}{3}$ + $\dfrac{1}{3^2}$  + ..... + $\dfrac{1}{3^{99}}$ (2)

Lấy (2) - (1) ta đc

$2N = 1 - \dfrac{1}{3^{100}} = \dfrac{3^{100} - 1}{3^{100}}$

$ => N = \dfrac{3^{100} - 1}{3^{100}} : 2 = \dfrac{3^{100} - 1}{3^{100}.2}$

d, Ta có : 

$D = 7 + 7^2 + 7^3 + .... + 7^{100} (1)$

$ => 7D = 7^2 + 7^3 + 7^4 + .... + 7^{101} (2)$

Lấy (2) - (1) ta đc

$6D = 7^{101} - 7$

$ => D = \dfrac{7^{101} - 7}{6}$

Giải thích các bước giải: