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\(A=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)(Dấu . là dấu nhân)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{9}{27}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\)
A = 4/3x7 + 4/7x11+ 4/11x15 + 4/15x19 + 4/19 x23 + 4/23 x 27
A = 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23 -1/27
A = 1/3 - 1/27
A = 8/27
Mình sửa lại đề một chút nhé.
\(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\).
A = \(\dfrac{4}{3\times7}\) + \(\dfrac{4}{7\times11}\) + \(\dfrac{4}{11\times15}\) + \(\dfrac{4}{15\times19}\) + \(\dfrac{4}{19\times23}\) + \(\dfrac{4}{23\times27}\)
A =1/3 -1/7+1/7-1/11 + 1/11-1/15 + 1/15 - 1/19 + 1/19 -1/23+1/23-1/27
A = 1/3 - 1/27
A = 8/27
\(C=\frac{4}{3x7}+\frac{4}{7x11}+\frac{4}{11x15}+\frac{4}{15x19}+\frac{4}{19x23}+\frac{4}{23x27}\)
= 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27
=1/3-(1/7+1/7)-(1/11+1/11)-(1/15-1/15)-(1/19+1/19)-(1/23-1/23)-1/27
=1/3-1/27
=...
=8/27
\(S=\frac{7-3}{3\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}\)
n: là số thứ tự của số hạng.
\(S=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}-\frac{1}{15}+...+\frac{1}{4n-1}-\frac{1}{4n+3}=\frac{1}{3}-\frac{1}{4n+3}\)
\(S=\frac{4n}{3\left(4n+3\right)}=\frac{664}{1995}\Leftrightarrow\frac{n}{4n+3}=\frac{166}{665}\Leftrightarrow665n=664n+3\cdot166\Leftrightarrow n=498\)
a) Vậy số hạng cuối cùng của dãy là: \(\frac{1}{\left(4\cdot498-1\right)\left(4\cdot498+3\right)}=\frac{1}{1991\cdot1995}\)
b) Tổng S có 498 số hạng.
Ta có : \(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x.\left(x+4\right)}=\frac{5}{63}\)
=> \(\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{5}{63}\)
=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{20}{63}\)
=> \(\frac{1}{3}-\frac{1}{x+1}=\frac{20}{63}\)
=> \(\frac{1}{x+1}=\frac{1}{63}\)
=> x + 1 = 63
=> x = 62
Vậy x = 62
Sửa lại bài làm của XYZ một chút:
=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+..+\)\(\frac{1}{x}-\frac{1}{x+4}\)
=> \(\frac{1}{3}-\frac{1}{x+4}\)= \(\frac{5}{63}\div\frac{1}{4}=\frac{20}{63}\)f
\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)
