Find the first six terms of the sequence whose first term is 1 and whose (n+l)th term is obtained by adding n to the nth term.

Solution

Let $straight t subscript straight n$ denote the nth term
               $straight t subscript 1 space end subscript space equals space 1$
According to given equation $straight t subscript straight n plus 1 end subscript space equals space straight t subscript straight n plus straight n$                         ...(i)
Putting n = 1 in (i), we get          $straight t subscript 1 plus 1 end subscript space equals space straight t subscript 1 plus 1 space equals space 1 plus 1 space equals space 2$       $rightwards double arrow space straight t subscript 2 space equals space 2$
Putting n = 2 in (i), we get          $straight t subscript 2 plus 1 end subscript space equals space straight t subscript 2 plus 2 space equals space 2 plus 2 space equals space 4$    $space space rightwards double arrow space space straight t subscript 3 space equals space 4$
Putting n = 3 in (i), we get          $straight t subscript 3 plus 1 end subscript space equals space straight t subscript 3 plus 3 space equals space 4 plus 3 equals 7$       $space space rightwards double arrow space space space space straight t subscript 4 space equals space 7$
Putting n = 4 in (i), we get         $straight t subscript 4 plus 1 end subscript space equals space straight t subscript 4 plus 4 space equals space 7 space plus space 4 space equals space 11$     $rightwards double arrow space straight t subscript 5 equals 11$
Putting n = 5 in (i), we get        $straight t subscript 5 plus 1 end subscript space equals space straight t subscript 5 space plus space 5 space equals space 11 space plus space 5 space equals space 16$    $rightwards double arrow space straight t subscript 6 space equals space 16$