Tính các tổng sau : A = 5 + $5^{2}$ + $5^{3}$ + . . . + $5^{99}$ + $5^{100}$ B = 1 + $4^{2}$ + $4^{4}$ + $4^{6}$ + . . . + $4^{300}$ C = 1 +

0 bình luận về “Tính các tổng sau : A = 5 + $5^{2}$ + $5^{3}$ + . . . + $5^{99}$ + $5^{100}$ B = 1 + $4^{2}$ + $4^{4}$ + $4^{6}$ + . . . + $4^{300}$ C = 1 +”

1. A = 5 + $5^{2}$ + $5^{3}$ +  . . . +$5^{99}$ + $5^{100}$

   ⇒ 5A = $5^{2}$ + $5^{3}$ + $5^{4}$ + . . . + $5^{100}$ + $5^{101}$

   ⇒ 5A – A = ( $5^{2}$ + $5^{3}$ + $5^{4}$ + . . . + $5^{100}$ + $5^{101}$ ) – ( 5 + $5^{2}$ + $5^{3}$ +  . . . +$5^{99}$ + $5^{100}$ )

   ⇒ 4A = $5^{101}$ – 5

   ⇒ A= $\frac{5^{101} – 5}{4}$ 

   ————-

   B = 1 + $4^{2}$ + $4^{4}$ +  . . . + $4^{300}$

   ⇒ $4^{2}$B = $4^{2}$ + $4^{4}$ + $4^{6}$ +  . . . + $4^{302}$

   ⇒ 16B = $4^{2}$ + $4^{4}$ + $4^{6}$ +  . . . + $4^{302}$

   ⇒ 16B – B = ( $4^{2}$ + $4^{4}$ + $4^{6}$ +  . . . + $4^{302}$ ) – ( 1 + $4^{2}$ + $4^{4}$ +  . . . + $4^{300}$ )

   ⇒ 15B = $4^{302}$ – 1

   ⇒ B =$\frac{4^{302} – 1}{15}$

   ————-

   C = 1 + $3^{2}$ + $3^{4}$ +  . . . + $3^{2020}$

   ⇒ $3^{2}$C = $3^{2}$ + $3^{4}$ + $3^{6}$ +  . . . + $3^{2022}$

   ⇒ 9C = $3^{2}$ + $3^{4}$ + $3^{6}$ +  . . . + $3^{2022}$

   ⇒ 9C – C = ( $3^{2}$ + $3^{4}$ + $3^{6}$ +  . . . + $3^{2022}$ ) – ( 1 + $3^{2}$ + $3^{4}$ +  . . . + $3^{2020}$ )

   ⇒ 8C = $3^{2022}$ – 1

   ⇒ C = $\frac{3^{2022} – 1}{8}$

   CHÚC BẠN HỌC TỐT <3

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2. Giải thích các bước giải:

   `A =5 + 5^2 + 5^3 +…+ 5^99 + 5^100`

   `=> 5A = 5^2 + 5^3 + 5^4 +…+ 5^100 + 5^101`

   `=> 5A – A = (5^2 + 5^3 + 5^4 +…+ 5^100 + 5^101)-(5 + 5^2 + 5^3 +…+ 5^99 + 5^100)`

   `=> 4A = 5^101 – 5`

   `=> A =(5^101 – 5)/4`

   __________

   `B = 1 + 4^2 + 4^4 + 4^6 +…+ 4^300`

   `=> 4^2 B = 4^2 + 4^6 + 4^8 +…+ 4^302`

   `=> 16B – B = (4^2 + 4^6 + 4^8 +…+ 4^302)-( 1 + 4^2 + 4^4 + 4^6 +…+ 4^300)`

   `=> 15B = 4^302 – 1`

   `=> B = (4^302 – 1)/15`

   __________

   `C = 1 + 3^2 + 3^4 +…+ 3^2020`

   `=> 3^2 C = 3^2 + 3^4 + 3^6 +….+ 3^2022`

   `=> 9C – C = (3^2 + 3^4 + 3^6 +….+ 3^2022)-(1 + 3^2 + 3^4 +…+ 3^2020)`

   `=> 8C = 3^2022 – 1`

   `=> C = (3^2022 – 1)/8`

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