Substituting n = 20, we obtain
a20 = 20\(\frac{(20-2)}{20+3}\)
= 20\(\frac{(18)}{23}\)
= \(\frac{360}{23}\)
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to
take on sth: | to begin to have a particular quality or appearance; to assume sth |
take sb on: | to employ sb; to engage sb to accept sb as one’s opponent in a game, contest or conflict |
take sb/sth on: | to decide to do sth; to allow sth/sb to enter e.g. a bus, plane or ship; to take sth/sb on board |