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1/ A= 1.(100-1)+2(100-2)+3.(100-3)+...+49.(100-51)+50.(100-50)
= 1.100-1+2.100 - 2.2 + 3.100 -3.3 + ...+49.100 - 49.51 + 50.100 - 50.50
= 100( 1 + 2 + 3 + ...+ 50) - ( 1 + 22 + 32 + ... + 502 )
= 127500- 42925
= 84575
2/ A= 1.3 + 5.7 + 9.11+ 13.15 + 17.19 + ... + 97. 101
= 1.3 + 5(6 + 1) +9( 6+ 5) + 13(6+9) + 17(6+13) + ... + 97(95+6)
= 3 + 5.6 + 1.5 + 9.6 + 5.9 + 13.6 + 9.13 + 17.6 + 13.17 + ... + 95.97 + 97.6
= 3 + ( 1.5 + 5.9 + 9.13 + 13.17 + ...+ 95.97) + 6( 5 + 9 + 13 + 17 + ... + 97)
= ...
=\(\frac{509447}{6}\)
Ta có:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}:2=\frac{25}{51}\)
Bài giải
\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\)
\(B=1\cdot2\cdot\left(3-1\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot\left(101-1\right)\)
\(B=1\cdot2\cdot3-1\cdot2+2\cdot3\cdot4-2\cdot3+...+99\cdot100\cdot101-99\cdot100\)
\(B=\left(1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\right)-\left(1\cdot2+2\cdot3+...+99\cdot100\right)\)
Đặt \(C=1\cdot2\cdot3+2\cdot3\cdot4+...+99\cdot100\cdot101\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+99\cdot100\cdot101\cdot\left(102-98\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+99\cdot100\cdot101\cdot102-98\cdot99\cdot100\cdot101\)
\(4C=99\cdot100\cdot101\cdot102\)
\(4C=101989800\)
\(C=101989800\text{ : }4\)
\(C=25497450\)
\(S_n=1.1!+2.2!+3.3!+...+n.n!\)
\(\text{Ta có:}\) \(1.1!=2!-1!\)
\(2.2!=3!-2!\)
\(3.3!=4!-3!\)
.......
\(n.n!=\left(n+1\right)!-n!\)
Cộng vế với vế ta đc:
\(S_n=1.1!+2.2!+3.3!+...+n.n!=2!-1!+3!-2!+4!-3!+...+\left(n+1\right)!-n!\)
\(=\left(n+1\right)!-1!=\left(n+1\right)!-1\)
