a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

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

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

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

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

a) \(3.5^2+15.2^2-26\div2\)

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) \(5^3.2-100\div4+2^3.5\)

= 125.2 - 25 + 8.5

= 250 - 25 + 40

= 225 + 40

= 265

c)\(6^2\div9+50.2-3^3.33\)

= 36 : 9 + 100 - 9.33

= 4 + 100 - 297

= 104 - 297

= -193

d)\(3^2.5+2^3.10-81\div3\)

= 9.5 + 8.10 - 27

= 45 + 80 - 27

= 125 - 27

= 98

e) \(5^{13}\div5^{10}-25.2^2\)

= 5³ - 25.4

= 125 - 100

= 25

f) \(20\div2^2+5^9\div5^8\)

= 20 : 4 + 5

= 5 + 5

= 10

Bài 1:

$B=1+3+3^2+3^3+...+3^{100}$

$=1+(3+3^2)+(3^3+3^4)+...+(3^{99}+3^{100})$

$=1+3(1+3)+3^3(1+3)+...+3^{99}(1+3)$

$=1+(1+3)(3+3^3+...+3^{99})=1+4(3+3^3+....+3^{99})$

$\Rightarrow B$ chia 4 dư 1.

Bài 2:

$C=5-5^2+5^3-5^4+...+5^{2023}-5^{2024}$

$5C=5^2-5^3+5^4-5^5+...+5^{2024}-5^{2025}$

$\Rightarrow C+5C=5-5^{2025}$

$6C=5-5^{2025}$

$C=\frac{5-5^{2025}}{6}$

a) \(S=1+2+2^2+..+2^{2022}\)

\(2S=2+2^2+2^3+...+2^{2023}\)

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

\(3S=3^2+3^3+...+3^{2023}\)

\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

 a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)

b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)

c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)

\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)

\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)

Tương tự câu d,e,f bạn tự làm nhé

trên đầu bài là giấu phẩy hay giấu nhân thế

\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)

\(b,3^2=9,3^3=27,3^4=81,3^5=243\)

\(c,4^2=16,4^3=64,4^4=256\)

\(d,5^2=25,5^3=125,5^4=625\)

a: \(2^2=4\)

\(2^3=8\)

\(2^4=16\)

\(2^5=32\)

\(2^6=64\)

\(2^7=128\)

\(2^8=256\)

\(2^9=512\)

\(2^{10}=1024\)

b: \(3^2=9\)

\(3^3=27\)

\(3^4=81\)

\(3^5=243\)

c: \(4^2=64\)

\(4^3=256\)

\(4^4=1024\)

d: \(5^2=25\)

\(5^3=125\)

\(5^4=625\)

siuu