The sequence N of natural numbers is divided into classes as follows
$1$
2 3 4
5 6 7 8 9
10 11 12 13 14 15 16
.......................................................
......................................................................
Find the sum of the numbers of n-th row

3 n^{3} + 4 n^{2} + 4 n

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3 n^{3} - 4 n^{2} + 4 n

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\frac{3 n^{3} - 4 n^{2} - 4 n}{2}

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\frac{3 n^{3} - 4 n^{2} + 4 n}{2}

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Solution

The correct option is D $\frac{3 n^{3} - 4 n^{2} + 4 n}{2}$
$W e c a n s e e t h a t a n y r o w h a s a n u m b e r w h i c h i s o n e m o r e t h a n t h e l a s t n a t u r a l n u m b e r u s e d i n p r e v i o u s r o w . l e t t h e b e g i n n i n g n u m b e r o f n^{t h} r o w i s A_{n} t h e n A_{1} = 1, A_{2} = 2, A_{3} = 5, A_{4} = 10 A_{2} - A_{1} = 1 A_{3} - A_{2} = 3 A_{4} - A_{3} = 5 ∴ A_{n} - A_{1} = 1 + 3 + 5 + . . . . . . u p t o (n - 1) t e r m s ∴ A_{n} = (n - 1)^{2} + A_{1} = (n - 1)^{2} + 1 ∴ s u m o f n^{t h} r o w = ((n - 1)^{2} + 1) + (n^{2} + 1) + . . . . . . . u p t o (2 n - 1) t e r m s = (n - 1)^{2} + n^{2} + . . . . . . (2 n - 1) t e r m s + (2 n - 1) = (\frac{3 n^{3} - 4 n^{2} + 4 n}{2})$

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