Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
$3S_n=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{(n+3)-n}{n(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{1.2.3.3}-\frac{1}{3(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}$
Giải tạm trong câu này chứ không thấy đề ở đâu hết. Với n dương
So sánh \(\frac{n}{n+3};\frac{n+1}{n+2}\)
Ta có: \(\frac{n}{n+3}< \frac{n}{n+2}\) (vì cùng tử nên mẫu bé hơn thì lớn hơn) (1)
Ta lại có: \(\frac{n}{n+2}< \frac{n+1}{n+2}\) (vì cùng mẫu nên tử lớn hơn thì lớn hơn) (2)
Từ (1) và (2) \(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)
\(S=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{20.21.22.23}\)
\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+......+\frac{3}{20.21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.....+\frac{1}{20.21.22}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}.\frac{295}{1771}=\frac{295}{3542}\)
F= \(\frac{1}{1.2.3}\)- \(\frac{1}{2.3.4}\)+ \(\frac{1}{2.3.4}\)- \(\frac{1}{3.4.5}\)+....+\(\frac{1}{47.48.49}\)- \(\frac{1}{48.49.50}\)
F=\(\frac{1}{1.2.3}\)- \(\frac{1}{48.49.50}\)
F=\(\frac{6533}{39200}\)
\(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\)= \(\frac{1}{3}\left(\frac{1}{n\cdot\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\right)\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\cdot\frac{6533}{39200}=\frac{6533}{117600}\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)