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a) 2 + 6 + 10 + 14 +...+202
= 2.1 + 2.3 + 2.5 + 2.7 +...+2.101
=2.(1+3+5+7+...+101)
=2.[(1+101).51:2]
=2.2601
=5202
b) Đặt A=1+2+22+23+...+265
=> 2A=2+22+23+24+...+266
=>2A-A=266-1
A=266-1
+) Số số hạng của dãy là : \(\left(202-2\right):4+1=51\) (số)
Tổng của dãy là : \(\frac{\left(202+2\right)\times51}{2}=5202\)
+) Đặt \(A=1+2+2^2+2^3+...+2^{65}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{66}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+...+2^{66}\right)-\left(1+2+2^2+2^3+...+2^{65}\right)\)
\(\Rightarrow A=2^{66}-1\)
+) Đặt \(B=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5B=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5B-B=4B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4B=5^{101}-5\)
\(\Rightarrow B=\frac{5^{101}-5}{4}\)
_Chúc bạn học tốt_
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
Bn tự tính típ nha
a/ \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+........+\frac{99}{100!}\)
\(\Leftrightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+......+\frac{100-1}{100!}\)
\(\Leftrightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+.....+\frac{100}{100!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{99!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{100!}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}\)
1)
Dễ thấy \(B=\dfrac{10^{19}}{10^{19}-3}>1\)
\(\Rightarrow B=\dfrac{10^{19}}{10^{19}-3}>\dfrac{10^{19}+2}{10^{19}-3+2}=\dfrac{10^{19}+2}{10^{19}-1}=A\)
bài 1 mifk viết sai nha.
bài 1: cho A=1+3+3\(^2\)+3\(^3\)+...+3\(^{10}\).Tìm số tự nhiên n biết 2 x A + 1 = 3\(^n\)
B1:
\(A=1+3+3^2+3^3+...+3^{10}\\ 3A=3+3^2+3^3+3^4+...+3^{11}\\ 3A-A=3^{11}-1\\ \Rightarrow A=\frac{3^{11}-1}{2}\)
mấy câu khác tương tự nha
a) \(1+2^1+2^2+2^3+....+2^{10}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+....+2^{10}+2^{11}\right)-\left(1+2+2^2+2^3+....+2^{10}\right)\)
\(\Rightarrow A=2^{11}-1\)
b) \(3+3^2+3^3+3^4+.....+3^{100}\)
\(3A=3^2+3^3+3^4+....+3^{100}+3^{101}\)
\(3A-A=\left(3^2+3^3+3^4+....+3^{100}+3^{101}\right)-\left(3+3^2+3^3+....+3^{100}\right)\)
\(2A=3^{101}-3\)
\(A=\frac{3^{101}-3}{2}\)
c) \(2+2^2+2^3+....+2^{100}\)
\(2A=2^2+2^3+2^4+....+2^{100}+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+....+2^{100}+2^{101}\right)-\left(2+2^2+2^3+.....+2^{100}\right)\)
\(A=2^{101}-2\)