ta có :

2.S=\(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\)

2.S-S=\(\left(\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\right)-\left(\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

=>S=\(\frac{3}{2^{10}}-\frac{3}{2}\)

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(=3\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)=3\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}\)

\(S=3\left(1+\frac{1}{2^{ }}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

\(2S=3\left(\frac{2}{2^0}+\frac{2}{2^1}+\frac{2}{2^2}+...+\frac{2}{2^9}\right)=3\left(2+1+\frac{1}{2^{ }}+...+\frac{1}{2^8}\right)\)\(2S-S=S=3\left(2+1+\frac{1}{2^1}+...+\frac{1}{2^8}\right)-3\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}\)

dsad221371áadsadsada

lộn 509/521 +3 =2072/521

Ta có: S = 3+3/2+3/2^2+3/2^3+...+3/2^9

    1/2.S = 3/2+3/2^2+3/2^3+3/2^4+...+3/2^10

\(\Rightarrow\) S-1/2.S = 3 - 3/2^10

\(\Rightarrow\) 1/2.S = 3 - 3/2^10

\(\Rightarrow\) S = (3 - 3/2^10) : 1/2 

\(\Rightarrow\) S = 6 - 6/2^10

Nếu đúng thì cho mk biết nha

Bạn Đức Nguyễn Ngọc làm đúng đó bạn

ban duc nguyen ngoc lam dung day

1/

\(N=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)=\)

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

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

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

\(N=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe

\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)

\(S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)

Đặt \(N=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\)

\(\Rightarrow2N-N=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)

\(\Rightarrow N=2-\frac{1}{2^9}\)

Khi đó \(S=3.N=3.\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}=\frac{3069}{512}\)

S=115,330078781

ket qua= 3069/512