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TH1)
\(B=\dfrac{2}{11x15}+\dfrac{2}{15x19}+\dfrac{2}{19x23}+......+\dfrac{2}{51x55}\)
\(B=\dfrac{2}{11}-\dfrac{2}{15}+\dfrac{2}{15}-\dfrac{2}{19}+\dfrac{2}{19}-\dfrac{2}{23}+.....+\dfrac{2}{51}-\dfrac{2}{55}\)
\(B=\dfrac{2}{11}-\dfrac{2}{55}\)
\(B=\dfrac{8}{55}\)
TH2)
\(B=\dfrac{2}{11x15}-\dfrac{2}{15x19}-\dfrac{2}{19x23}-......-\dfrac{2}{51x55}\)
\(B=\dfrac{2}{11}+\dfrac{2}{15}-\dfrac{2}{15}+\dfrac{2}{19}-\dfrac{2}{19}+\dfrac{2}{23}-....-\dfrac{2}{51}+\dfrac{2}{55}\)
\(B=\dfrac{2}{11}+\dfrac{2}{55}\)
\(B=\dfrac{12}{55}\)
ta có:
A=2/4(4/11.15+4/15.19+4/19.23+.....+4/51.55)
A=2/4(1/11-1/15+1/15-1/19+1/19-1/23+....+1/51-1/55)
A=2/4(1/11-1/55)
A=2/4*4/55=8/220=2/55
B=-55/3/*8/3=-165/24=-55/8
suy ra A*B=2/55*(-55/8)=-1/4
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
suy ra:
2A= 2 +2^2+ 2^3 + 2^4 + 2^5+ 2^6+ 2^7
suy ra
2A-A= 1+2^7
còn mấy câu còn lại tương tự thui bạn ak
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
\(C=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.13}+...+\frac{2}{51.55}\)
\(2.C=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)
\(2.C=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)
\(2.C=\frac{1}{11}-\frac{1}{55}\)
\(2.C=\frac{5}{55}-\frac{1}{55}=\frac{4}{55}\)
\(C=\frac{4}{55}:2=\frac{4}{55}.\frac{1}{2}=\frac{2}{55}\)
Vậy \(C=\frac{2}{55}\)
đọc kĩ đi bạn ơi
có cả số 1 nữa