\(A=1+3^2+3^4+...+3^{98}+3^{100}\)

\(3^2\cdot A=3^2+3^4+3^6+...+3^{100}+3^{102}\)

\(9A-A=\left(3^2+3^4+3^6+...+3^{100}+3^{102}\right)-\left(1+3^2+3^4+...+3^{98}+3^{100}\right)\)

\(8A=3^{102}-1\)

\(\Rightarrow A=\dfrac{3^{102}-1}{8}\)

A = 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰
3A = 3. ( 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰ )
3A = 3. 1 + 3. 3² + 3. 3⁴ + ..... + 3. 3⁹⁸ + 3. 3¹⁰⁰
3A = 3² + 3³ + 3⁴ + ..... + 3¹⁰⁰ + 3¹⁰¹
3A - A = ( 3² + 3³ + 3⁴ + ..... + 3¹⁰⁰ + 3¹⁰¹ ) - ( 1 + 3² + 3⁴ + ..... + 3⁹⁸ + 3¹⁰⁰ )
2A = 3¹⁰¹ - 1
A = ( 3¹⁰¹ - 1 ) : 2

Lời giải:

$A=1+32+34+....+398+400$

Từ $32$ đến $400$ có số số hạng là:

$(400-32):2+1=185$ (số hạng)

$32+34+....+398+400=(400+32).185:2=39960$

$\Rightarrow A=1+39960=39961$

Tham khảo

Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)

3A = 3+32+33+...+3100+31013+32+33+...+3100+3101

Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)

2A = 3101−13101−1

⇒⇒ A = 3101−123101−12

Vậy A = 3101−12

A=3 mũ 101-1 phân số2

\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)

       A =  1 - 3 +  3² -   3³ + 3⁴ - ... + 3⁹⁸ - 3⁹⁹ + 3¹⁰⁰

      3A =  3 - 3² + 3³ - 3⁴+ 3⁵  - ... + 3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹

3A + A = 3 - 3²+ 3³-3⁴+3⁵ -...+3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹ + 1 - 3 +...-3⁹⁹+3¹⁰⁰

4A   =    3¹⁰¹ + 1

  A    = \(\dfrac{3^{101}+1}{4}\) 

Theo đề bài ra, ta có :

`A=1+32+34+36+....+32008`

\(\Rightarrow\) `9A = 3^2 + 3^4 + 3^6 + 3^8 + ... + 3^2010`

`9A - A=(32+34+36+38+....+ 32010)-(1+32+34+36+....+ 32008)`

\(\Rightarrow\) `8A=(-1)+32010`

\(\Rightarrow\) `8A-32010=(-1)`

@Nae

Theo đề bài ra, ta có :

A=1+3^2+3^4+3^6+....+3^2008

 9A = 3^2 + 3^4 + 3^6 + 3^8 + ... + 3^2010

9A - A= (3^2+3^4+3^6+3^8+....+ 3^2010)- (1+3^2+3^4+3^6+....+ 3^2008)

8A = -1+3^2010

 8A - 3^2010 = (-1)

@Nae

xin lỗi bài trên của mình làm sai

Ta có: 3A = 3.(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰⁾ 

3A = 3+3²+3³+...+3¹⁰⁰+3¹⁰¹

Suy ra: 3A – A = (3+3²+3³+...+3¹⁰⁰+3¹⁰¹)−(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰)

2A = 3¹⁰¹−1

⇒ A = 3¹⁰¹−1

             2               

Vậy A = 3¹⁰¹−1

                 2           

em den lam

\(A=1+3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{101}\)

Trừ theo vế:

\(\Rightarrow3A-A=\left(3+3^2+3^3+...3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-1\Rightarrow A=\dfrac{3^{101}-1}{2}\)

A =1+3+3² +3³ +...+ 3¹⁰⁰

3A=3.(3⁰+3+3² +3³ +...+ 3¹⁰⁰)

3A=3¹+3² +3³ +...+ 3¹⁰¹

3A-A=(3¹+3² +3³ +...+ 3¹⁰¹)-(3⁰+3+3² +3³ +...+ 3¹⁰⁰)

2A=3¹⁰¹-3⁰

A=(3¹⁰¹-1) :2

vậy A=(3¹⁰¹-1) :2

t.i.c cho mình nha