Ta có: \(1\cdot1+2\cdot2+\cdots+2021\cdot2021\)

\(=1^2+2^2+\cdots+2021^2\)

\(=2021\cdot\frac{\left(2021+1\right)\left(2\cdot2021+1\right)}{6}=2021\cdot2022\cdot\frac{4043}{6}\)

=2753594311

Đặt :

\(A=1+2+2^2+2^3+...............+2^{10}\)

\(\Leftrightarrow2A=2+2^2+2^3+..............+2^{11}\)

\(\Leftrightarrow2A-A=\left(2+2^2+2^3+.......+2^{11}\right)-\left(1+2+2^2+......+2^{10}\right)\)

\(\Leftrightarrow A=2^{11}-1\)

Đặt \(A=1+2+2^2+2^3+...+2^{10}\)

\(\Leftrightarrow2A=2+2^2+2^3+2^4+...+2^{11}\)

\(\Leftrightarrow A=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Leftrightarrow A=2^{11}-1\)

a)9^6

b)2^5

c)5^5

d)a^7

5

 Ta có : \(n.n!=\left(n+1-1\right).n!=n!.\left(n+1\right)-n!=\left(n+1\right)!-n!\)

Áp dụng : \(1.1!+2.2!+3.3!+4.4!+5.5!=2!-1!+3!-2!+4!-3!+5!-4!+6!-5!\)

\(=6!-1=1.2.3.4.5.6-1=719\)

Chị Ngọc giúp e bài toán đó nhé, thanks chị nhiều ạ

Ý gì vậy

S = 1.1! + 2.2! + 3.3! + 4.4! + 5.5! = 1 + 2.1.2 + 3.1.2.3 + 4.1.2.3.4 + 5.1.2.3.4.5 = 1 + 4 + 18 + 96 + 600 = 719

Đáp án :719

100% chính xác !!

Bài 2:

100! = 1.2.3.4.   ...................   .100

ta có công thức quy lạp \(1.1!+2.2!+...+n.n!=\left(n+1\right)!-1\)

áp dụng vào bài \(1.1!+2.2!+3.3!+...+7.7!=\left(7+1\right)!-1=8!-1=40320-1=40319\)

Thấy :            \(\frac{1}{1.1!}=\frac{1}{1}\)

                       \(\frac{1}{2.2!}=\frac{1}{4}\)

                       \(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )

                         ...

                       \(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)

Cộng từng vế có :

  \(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)

Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)