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Sum of n terms of sequence

1+2+4+7+11+16+....... is

AND please explain the methord

$T h e g i v e n s e q u e n c e i s, 1 + 2 + 4 + 7 + 11 + 16 + ..... H e r e, 1, 2, 3, 4, .... i s t h e s u c c e s s i v e d i f f e r e n c e, w h i c h a r e i n A . P . N o w, S = 1 + 2 + 4 + 7 + 11 + 16 + ..... + T_{n} S = 1 + 2 + 4 + 7 + 11 + 16 + ..... + T_{n - 1} + T_{n} O n s u b t r a c t i n g, w e g e t 0 = 1 + 1 + 2 + 3 + 4 + 5 + ........ (n - 1) t e r m s - T_{n} \Rightarrow T_{n} = 1 + (1 + 2 + 3 + 4 + 5 + ........ (n - 1) t e r m s) \Rightarrow T_{n} = 1 + [\frac{n - 1}{2} (2 \times 1 + (n - 1 - 1) \times 1)] \Rightarrow T_{n} = 1 + [\frac{n (n - 1)}{2}] T h u s, S_{n} = Σ T_{n} = Σ 1 + [\frac{n (n - 1)}{2}] = Σ (1) + \frac{1}{2} Σ n^{2} - \frac{1}{2} Σ n = n + \frac{1}{2} (\frac{n (n + 1) (2 n + 1)}{6}) - \frac{1}{2} (\frac{n (n + 1)}{2}) = \frac{12 n + n (n + 1) (2 n + 1) - 3 n (n + 1)}{12} = \frac{12 n + n (n + 1) (2 n - 2)}{12} H e n c e, s u m o f n t e r m s o f g i v e n s e q u e n c e i s \frac{12 n + n (n + 1) (2 n - 2)}{12} .$

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