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\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{16}\left(1+2+3+....+16\right)\)
\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+....+\frac{1}{16}\cdot\frac{16.17}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{17}{2}\)
\(A=\frac{\left(2+3+4+....+17\right)}{2}=\frac{\left(2+17\right).\left(17-2+1\right):2}{2}=\frac{152}{2}=76\)
Ta có: \(A = 1+{1+2\over 2} + {1+2+3\over 3} +...+{1+2+...+ 16\over 16}\)
Xét: \(S_n = 1+2+3+...+n =\frac{n(n+1)}{n} (n \in N^*)\)
=> \({S_n\over n} = {(n+1)\over 2}\)
Thay vào biểu thức A ta có:
\(A=1 + {3\over 2} + {4\over 2} + ... + {17\over 2}\)
\(A={(2+3+4+...+17)\over 2}\)
\(A={(17+2)[(17-2+1):2]\over 2} = {152\over2}=76\)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)
a) \(\left(\frac{11}{12}:\frac{44}{16}\right).\left(\frac{-1}{3}+\frac{1}{2}\right)\) \(=\left(\frac{11}{12}.\frac{16}{44}\right).\left(\frac{-2}{6}+\frac{3}{6}\right)\) \(=\frac{1}{3}.\frac{1}{6}\) \(=\frac{1}{18}\)
b) \(\frac{\left(-5\right)^2.\left(-5\right)^3.16}{5^4.\left(-2\right)^4}\) \(=\frac{\left(-5\right)^5.2^4}{5^4.\left(-2\right)^4}\) \(=5\) (Có sửa đề lại, nếu có sai thì ib mình sửa lại nhé!)
c) \(7,5:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(\frac{-5}{3}\right)\) \(=\frac{15}{2}.\left(\frac{-3}{5}\right)+\frac{5}{2}.\left(\frac{-3}{5}\right)\) \(=\frac{-3}{5}.\left(\frac{15}{2}+\frac{5}{2}\right)\)
\(=\frac{-3}{5}.10\) \(=-6\)
d) \(\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right):\frac{5}{4}\) \(=\left(\frac{-1}{2}+\frac{1}{3}\right).\frac{4}{5}+\left(\frac{2}{3}+\frac{1}{2}\right).\frac{4}{5}\)
\(=\frac{4}{5}.\left(\frac{-1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{2}\right)\) \(=\frac{4}{5}.\left(\frac{0}{2}+1\right)\) \(=\frac{4}{5}.1=\frac{4}{5}\)
a) (1112:4416).(−13+12)(1112:4416).(−13+12) =(1112.1644).(−26+36)=(1112.1644).(−26+36) =13.16=13.16 =118=118
b) (−5)2.(−5)3.1654.(−2)4(−5)2.(−5)3.1654.(−2)4 =(−5)5.2454.(−2)4=(−5)5.2454.(−2)4 =5=
c) 7,5:(−53)+212:(−53)7,5:(−53)+212:(−53) =152.(−35)+52.(−35)=152.(−35)+52.(−35) =−35.(152+52)=−35.(152+52)
=−35.10=−35.10 =−6=−6
d) (−12+13).45+(23+12):54(−12+13).45+(23+12):54 =(−12+13).45+(23+12).45=(−12+13).45+(23+12).45
=45.(−12+13+23+12)=45.(−12+13+23+12) =45.(02+1)=45.(02+1) =45.1=45
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+...+16\right)\)
\(A=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.120\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{15}{2}=\frac{2+3+4+...+15}{2}=\frac{119}{2}=59,5\)
Vậy \(A=59,5\)
A=1+\(\frac{1+2}{2}\)+\(\frac{1+2+3}{3}\)+...+\(\frac{1+2+3+...+16}{16}\)
A=\(\frac{2}{2}\)+\(\frac{3}{2}\)+\(\frac{4}{2}\)+...+\(\frac{17}{2}\)
A=\(\frac{2+3+4+...+17}{2}\)
A=76(đề thi HSG huyện tui có tui làm zậy mà cũng có điểm tuyệt đối)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+3+....+16\right)\)
\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+\frac{1}{4}\cdot\frac{4.5}{2}+.....+\frac{1}{16}\cdot\frac{16.17}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}\)
\(A=\frac{\left(2+3+4+.....+17\right)}{2}=\frac{\left(2+17\right).16}{2}=\frac{152}{2}=76\)