Cho \(S=1\cdot2\cdot3+2\cdot3\cdot4+.....+n\left(n+1\right)\left(n+2\right)\) Với \(n\inℕ^∗\) CM
4S +1 là số chính phương
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M
13 tháng 6 2020
A = 1.2.3 + 2.3.4 + 3.4.5 ... + n(n + 1)(n + 2)
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + n(n + 1)(n + 2).4
4A = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2)+ ... + n(n + 1)(n + 2)[(n + 3) - (n - 1)]
4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + n(n + 1)(n + 2)(n + 3) - (n-1)n(n+1)(n+2)
4A = n(n+1)(n+2)(n+3)
A = n(n + 1)(n+2)(n + 3) : 4
7 tháng 2 2017
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
27 tháng 12 2018
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)
=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)
\(S=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)
\(4S=1.2.3.4+2.3.4.4+...+n\left(n+1\right)\left(n+2\right).4\)
\(4S=1.2.3.4+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\)
\(\left[\left(n+3\right)-\left(n-1\right)\right]\)
\(4S=1.2.3.4+2.3.4.5-1.2.3.4+...+\)
\(n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
\(4S=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(4S+1=n\left(n+3\right)\left(n+1\right)\left(n+2\right)+1\)
\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)
Đặt \(n^2+3n=t\)
\(Đt=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)(là số chính phương)