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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 = \(\dfrac{2^{101}-1}{3.2^{100}}\)
Đặt \(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.......\frac{899}{30^2}\)
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}........\frac{29.31}{30^2}\)
\(A=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).....\left(29.31\right)}{2^2.3^2.4^2.....30^2}\)
Để dễ rút gọn,ta viêt tử số của A dưới dạng tích các số tự nhiên liên tiếp:
\(A=\frac{\left(1.2.3.......29.29\right).\left(3.4.5.........30.31\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)........\left(30.30\right)}\)
\(A=\frac{1.2.3.......28.29}{2.3.4.........29.30}.\frac{3.4.5.....30.31}{2.3.4.......29.30}=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
Vậy A=31/60
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Sai rồi, để tôi sửa lại:
\(\frac{3}{2^2}.\frac{8}{3^2}.....\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{29.31}{30.30}\)
\(=\frac{1.2.3.....31}{2.3.4.....30}.\frac{3.4.5....29}{2.3.4....30}=31.\frac{1}{60}=\frac{31}{60}\)