Skip to main content Home Courses Tests Daily Sign in Calculate: 1 1 ⋅ 2 + 1 2 ⋅ 3 + 1 3 ⋅ 4 + ⋯ + 1 9 ⋅ 10 \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{9\cdot10} 1 ⋅ 2 1 + 2 ⋅ 3 1 + 3 ⋅ 4 1 + ⋯ + 9 ⋅ 10 1 A 1 10 \frac{1}{10} 10 1 B 8 9 \frac{8}{9} 9 8 C 1 1 1 D ✓ 9 10 \frac{9}{10} 10 9 E 10 11 \frac{10}{11} 11 10 WORKED EXAMPLE
Solution Write each term as a difference: 1 n ( n + 1 ) = 1 n − 1 n + 1 \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} n ( n + 1 ) 1 = n 1 − n + 1 1 . The sum then becomes ( 1 − 1 2 ) + ( 1 2 − 1 3 ) + ⋯ + ( 1 9 − 1 10 ) \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\dots+\left(\frac{1}{9}-\frac{1}{10}\right) ( 1 − 2 1 ) + ( 2 1 − 3 1 ) + ⋯ + ( 9 1 − 10 1 ) , and the intermediate terms cancel. This leaves: 1 − 1 10 = 9 10 1-\frac{1}{10}=\frac{9}{10} 1 − 10 1 = 10 9 . Solve today's problem