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\(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\)= \(\frac{1}{3}\left(\frac{1}{n\cdot\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)}\right)\)
Đặt \(A=1.2.3.4+2.3.4.5+...+97.98.99.100\)
\(5A=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5\)
\(5A=1.2.3.4.5+2.3.4.5.\left(6-1\right)+...+97.98.99.100.\left(101-96\right)\)
\(5A=1.2.3.4.5+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100\)
\(5A=97.98.99.100.101\)
\(A=\frac{97.98.99.100.101}{5}=1901009880\)
\(S=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{20.21.22.23}\)
\(=\frac{1}{3}\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+......+\frac{3}{20.21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.....+\frac{1}{20.21.22}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2.3}-\frac{1}{21.22.23}\right)\)
\(=\frac{1}{2}.\frac{295}{1771}=\frac{295}{3542}\)
Đặt S=1.2.3.4+2.3.4.5+...+97.98.99.100
5S=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5
5S=1.2.3.4.(5 - 0)+2.3.4.5.(6 - 1)+...+97.98.99.100.(101 - 96)
5S=1.2.3.4.5-0.1.2.3.4+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99
5S=97.98.99.100.101
S=97.98.99.20.101
=>S=1901009880
Đặt A = 1.2.3.4 + 2.3.4.5 + ... + 97.98.99.100
5A = 1.2.3.4.5 + 2.3.4.5.5 + ... + 97.98.99.100.5
5A = 1.2.3.4.5 + 2.3.4.( 6 - 1 ) + ... + 97.98.99.100.( 101 - 96 )
5A = 1.2.3.4.5 + 2.3.4.5.6 - 1.2.3.4.5 + ... + 97.98.99.100.101 - 96.97.98.99.100
5A = 97.98.99.100.101
A = 97.98.99.100.101 : 5
A = 97.98.20.101
A = 19202120