Question

find the sum of n terms of the series (4-n/1)+(4-2/n)+(4-3/n) +.......

Answer 1

$[tex](4 - 1 \div n) + (4 - 2 \div n)..... \\ 4 + 4 + 4.......... + 4 - (1 \div n + 2 \div n + 3 \div n........) \\ 4n + (n(n + 1) \div 2 \div n \\ 4n + \frac{n + 1}{2} \\ \frac{9n +1 }{2}$
[/tex]