Lời giải:
$3S_n=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{(n+3)-n}{n(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{1.2.3.3}-\frac{1}{3(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}$
Tính nhanh \(B=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)
Tính tổng S=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
mk là KHÁNH LINH.
Tính tổng :
\(S_n=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2_n+1}{\orbr{\begin{cases}\\n\left(n+1\right)^2\end{cases}}}\)
\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(F=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{47.48.49.50}\)
Tính
\(C=1+\frac{1}{\left(-3\right)}+\frac{1}{\left(-3\right)^2}+....+\frac{1}{\left(-3\right)^{2015}}\)
tính \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n}-1\right)\)Biết nEN;n>2
Chứng minh:\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)....\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
n thuộc n sao
Chứng minh : ( n thuộc N*)
\(\left(1-\frac{2}{6}\right)x\left(1-\frac{2}{12}\right)x\left(1-\frac{2}{20}\right)x...x\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
Tính S=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{n}\left(1+2+...+n\right)\)
Chứng minh :
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
