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Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
\(B=1+\frac{1}{2}+\frac{1}{2}^2+\frac{1}{2}^3+...+\frac{1}{2}^{100}\)
\(B=1+\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+...+\frac{1^{100}}{2^{100}}\)
\(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2B-B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{100}}\)
\(B=2-\frac{1}{2^{100}}=\frac{2^{99}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{99}-1}{2^{100}}\)
a, A = 1 + 3 + 3\(^{^2}\) + .... + 3\(^{100}\)
3A = 3 + 3\(^2\) + ..... + 3\(^{101}\)
Lấy 3A - A
\(\Rightarrow\) 2A = 3\(^{101}\) - 1
A = \(\frac{3^{101}-1}{2}\)
b, Áp dụng kiến thức câu a
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
=>\(A=1-\frac{1}{2^{100}}\)
a/ta gọi biểu thức trên là A.
ta có: A=1+2+22+...+2100
2A= 2x(1+2+22+...+2100)
2A= 2x1+2x2+22x2+...+2100x2
2A= 2+22+23+....+2101
2A-A=A=(2+22+23+....+2101)-(1+2+22+...+2100)
A= 2101-1
b/ làm tương tụ như câu a nhưng cuối cùng phải thêm '':2'' (vì lúc đó ta tính ra 3A - A =2A nên phải chia 2)
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
1/2.A=1/22+1/23+...+1/2101
=>1/2A-A=1/2101-1/2
=>-1/2A=1/2101-1/2
A=(1/2101-1/2):(-1/2)=(1/2101-1/2).(-2)
=1-1/2100
-Quy luật: Nhân mỗi vế của đẳng thức cho số thích hợp để tạo ra đẳng thức mới, khi cộng (hoặc trừ) mỗi vế của mỗi đẳng thức thì sẽ rút gọn bớt.
a) \(A=2-2^2+2^3-2^4+...+2^{99}-2^{100}\)
\(\Rightarrow2A=2^2-2^3+2^4-2^5+...+2^{100}-2^{101}\)
\(\Rightarrow2A+A=2^2-2^3+2^4-2^5+...+2^{100}-2^{101}+\left(2-2^2+2^3-2^4+...+2^{99}-2^{100}\right)\)
\(\Rightarrow A=-2^{101}+2\)
b,c) làm tương tự.
d) \(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow3D=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D-D=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow2D=3+\dfrac{1}{3^{100}}\)
\(\Rightarrow2D=\dfrac{3^{101}+1}{3^{100}}\Rightarrow D=\dfrac{3^{101}+1}{2.3^{100}}\)
e) làm tương tự nhưng đổi thành cộng.
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^
A= \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
2A= \(2.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
⇒ 2A- A= \(1-\dfrac{1}{2^{100}}\)
⇒ A= \(1-\dfrac{1}{2^{100}}\)
B= \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
3B= \(3.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)
3B= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
⇒ 3B- B= \(1-\dfrac{1}{3^{100}}\)
⇒ B.(3-1)= \(1-\dfrac{1}{3^{100}}\)
⇒ 2B= \(1-\dfrac{1}{3^{99}}\)
⇒ B= \(\left(1-\dfrac{1}{3^{99}}\right):2\)
⇒ B= \(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)