tính tổng: S=3+3/2+3/22+....+3/29
Tính tổng S :
S=3/2+3/2^2+...+3/2^9
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=\(\frac{3}{2^{10}}-\frac{3}{2}\)
Tính tổng ; S=3+ 3/2+3/2^2+...+3/2^9
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(=3\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)=3\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}\)
Tính tổng S=3+3/2+3/2^2+...+3/2^9
\(S=3\left(1+\frac{1}{2^{ }}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(2S=3\left(\frac{2}{2^0}+\frac{2}{2^1}+\frac{2}{2^2}+...+\frac{2}{2^9}\right)=3\left(2+1+\frac{1}{2^{ }}+...+\frac{1}{2^8}\right)\)\(2S-S=S=3\left(2+1+\frac{1}{2^1}+...+\frac{1}{2^8}\right)-3\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}\)
tính tổng:
S=3+3/2+3/2^2+3/2^3+.....+3/2^9
Tính tổng :
S = 3 + 3/2 + 3/2^2 + ... + 3/2^9
S=3+32+322+323+...+329" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_SVG">
Tính tổng S=3+3/2+3/2^2+3/2^3+...3/2^9
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
ban duc nguyen ngoc lam dung day
bai 1 :tính tổng N=1^2+2^2+3^2+...+99^2
bài2: tính tổng A=1+4+9+16+25+36+...+100000
bài3: tính tổng S=1^2+3^2+5^2+...+49^2
bài4:tính tổng S=1^2+3^2+5^2+...+99^2
giúp mik với mik đang cần gấp
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
Tính tổng S=3+3/2+3/22+...+3/29
\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)
\(S=3.\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)
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 \(S=3.N=3.\left(2-\frac{1}{2^9}\right)=6-\frac{3}{2^9}=\frac{3069}{512}\)
Tính tổng S=3+3\2+3\22+...+3\29
Tính tổng: S=3+3/2+3/22+...+3/29