\(A=1^2+2^2+3^2+...+n^2=\)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+n\left[\left(n+1\right)-1\right]=\)

\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n=\)

\(=\left[1.2+2.3+3.4+...+n\left(n+1\right)\right]-\left(1+2+3+...+n\right)=\)

Đặt 

\(B=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3B=1.2.3+2.3.3++3.4.3+n\left(n+1\right).3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)=\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}-\dfrac{n\left(n+1\right)}{2}\) Là 1 đa thức bậc 3

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

\(\Rightarrow\dfrac{1}{2}S=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)

trừ vế ta được :

\(S-\dfrac{1}{2}S=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)

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

\(\Rightarrow S=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{2^{101}}\right)\)

\(\Rightarrow S=1-\dfrac{1}{2^{100}}< 1\left(đpcm\right)\)

S=3+3^2+3^3+...+3^2022

3S=3.(3+3^2+3^3+...+3^2022)

3S=3^2+3^3+3^4+...+3^2023

⇒3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

⇒2S=3^2023-3

⇒S=3^2023-3 / 2

S=3+3^2+3^3+...+3^2022

=>3S=3^2+3^3+3^4+...+3^2023

=>3S-S=(3^2+3^3+3^4+...+3^2023)-(3+3^2+3^3+...+3^2022)

=>2S=3^2023-3

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

Vậy S=​\(\dfrac{3^{2023}-3}{2}\)​

\(\Rightarrow3S=3^2+3^3+3^4+...+3^{2023}\)

trừ vế với vế ta được :

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

\(\Rightarrow2S=3^{2023}-3\)

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

a.

$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}$

$2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}$

$\Rightarrow 2S-S=1-\frac{1}{2^{50}}$

$\Rightarrow S=1-\frac{1}{2^{50}}$

b.

$S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{20}}$

$3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{19}}$

$\Rightarrow 3S-S=1-\frac{1}{3^{20}}$

$\Rightarrow 2A=1-\frac{1}{3^{20}}$

$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{20}}$

@Trần Bảo Việt, bạn không trả lời linh tinh nhé!

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

\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{2022}+2^{2023}\)

trừ vế với vế ta được : 

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

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

A=1.3.(5-2)+3.5.(7-2)+5.7.(9-2)+...+97.99(101-2)=

=(1.3.5+3.5.7+5.7.9+...+97.99.101) - 2.(1.3+3.5+5.7+...+97.99)

Đặt 

B=1.3.5+3.5.7+5.7.9+...+97.99.101

=> 8B=1.3.5.8+3.5.7.8+5.7.9.8+...+97.99.101.8=

=1.3.5(7+1)+3.5.7.(9-1)+5.7.9.(11-3)+...+97.99.101.(103-95)=

=1.3.5+1.3.5.7-1.3.5.7+3.5.7.9-3.5.7.9+5.7.9.11-...-95.97.99.101+97.99.101.103=

=3.5+97.99.101.103

\(\Rightarrow B=\dfrac{15+97.99.101.103}{8}\)

Đặt 

C=1.3+3.5+5.7+...+97.99

=> 6C=1.3.6+3.5.6+5.7.6+...+97.99.6=

=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+97.99.(101-95)=

=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-95.97.99+97.99.101=

=1.3+97.99.101

\(\Rightarrow C=\dfrac{3+97.99.101}{6}\)

\(\Rightarrow A=B-2C\)

Thay các giá trị của B và C bạn tự tính nốt nhé

Nếu mua hai gói thì giá của mỗi gói là:

\(90000:2=45000\left(đồng\right)\)

Nếu mua ba gói thì giá của mỗi gói là:

\(130000:3=43333\left(đồng\right)\)

Mà \(50000>45000>43333\) nên mẹ khuyên Mai nên mua 3 gói là rẻ nhất là đúng.

Nếu mua hai gói thì giá của mỗi gói là: 90000/2=45000
 đồng

Nếu mua ba gói thì giá của mỗi gói là: 130000/3 đồng

Vì 130000/3< 45 000 < 50 000 đồng nên mua ba gói là rẻ nhất

Bài 1:

$A=1.2+2.3+3.4+...+201.202$

$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+201.202(203-200)$

$=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+201.202.203-200.201.202$

$=(1.2.3+2.3.4+3.4.5+...+201.202.203)-(1.2.3+2.3.4+....+200.201.202)$

$=201.202.203$

$\Rightarrow A=\frac{201.202.203}{3}=2747402$

Bài 2:

$S=4.5+5.6+6.7+....+100.101$

$3S=4.5(6-3)+5.6.(7-4)+6.7.(8-5)+....+100.101(102-99)$

$=4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+....+100.101.102-99.100.101$

$=(4.5.6+5.6.7+6.7.8+...+100.101.102)-(3.4.5+4.5.6+5.6.7+...+99.100.101)$

$=100.101.102-3.4.5$

$\Rightarrow S=\frac{100.101.102-3.4.5}{3}=343380$

\(a.\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\\ =\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\\ b.\dfrac{7}{12}-\dfrac{27}{7}\cdot\dfrac{1}{18}\\ =\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{31}{84}\\ c.\left(\dfrac{23}{11}-\dfrac{15}{82}\right)\cdot\dfrac{41}{25}\\ =\dfrac{1721}{902}\cdot\dfrac{41}{25}=\dfrac{1721}{550}\\ d.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\\ =\dfrac{13}{10}\cdot\left(\dfrac{-5}{13}\right)=\dfrac{-1}{2}\)