MCQ
$1 + (1 + 2) + (1 + 2 + 3) + …..n $ પદ સુધી = …..
- A$n^2 - 2n + 6$
- B$\frac{{n(n + 1)(2n - 1)}}{6}$
- C$n^2 + 2n + 6$
- D$\frac{{n(n + 1)(2n + 1)}}{6}$
$ = \sum\limits_{n = 1}^n {\,\sum\limits_{n = 1}^n n } \,\,\, = \sum\limits_{n = 1}^n {\,\,\frac{n}{2}(n + 1)} \,\,\, = \frac{1}{2}[\sum {{n^2} + \sum n } ]\,\,\,$
$ = \frac{1}{2}\left[ {\frac{n}{6}(n + 1)(2n + 1) + \frac{n}{2}(n + 1)} \right]$
$ = \frac{1}{2}\,.\,\frac{n}{6}(n + 1)(2n + 1 + 3)\,\,\, = \frac{{n(n + 1)(n + 2)}}{6}$
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