Find a formula for the general term of the squence 1, 2, 2, 3, 3, 3, 4, 4, 4,4, ...


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Question Number 55359 by gunawan last updated on 22/Feb/19

$Find a formula for the general$ $term of the squence$ $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Feb/19

	$1 \to n o s o f t e r m \to 1$ $(2, 2) \to d o \to 2$ $(3, 3, 3) \to d o \to 3$ $(4, 4, 4, 4) \to d o \to 4$ $. . .$ $. . .$ $(n, n, n, . . . n) \to d o \to n$ $n o w$ $1, 2, 2, 3, 3, 3, 4, 4, 4, 4 . . . . . n, n, n . . . . n$ $t o t a l n o s o f t e r m o f t h i s s e r i e s$ $s = 1 + 2 + 3 + 4 + . . + n$ $s = \frac{n}{2} [2 \times 1 + (n - 1) \times 1] \to \frac{n^{2} + n}{2}$ $e x a m p l e$ $l e t t o f i n d 19 t h t e r m o f t h e s e q u e n c e$ $p u t n = 5 i n \frac{n^{2} + n}{2} \to \frac{5^{2} + 5}{2} = 15$ $p u t n = 6 i n \frac{n^{2} + n}{2} \to \frac{6^{2} + 6}{2} = 21$ $21 > 19 > 15 s o T_{19} = 6$ $T_{100} 100 t h t e r m . . .$ $i) c h o s e n s u c h t h a t \frac{n^{2} + n}{2} n e a r e s t t o 100$ $\frac{14^{2} + 14}{2} = 105 \frac{13^{2} + 13}{2} = 91 \frac{15^{2} + 15}{2} = 120$ $1, (2, 2), (3, 3, 3), (4, 4, 4, 4), . . ., (13, 13, . . 13)$ $i f w e c o u n t n o s o f t e r m s i t i s 91$ $1, (2, 2), (3, 3, 3), (4, 4, 4, 4) . . . (14, 14, 14, . . 14)$ $i f w e c o u n t n o s o f t e r m s = 91 + 14 = 105$ $s o T_{100} = 14$ $s o T_{r} = n [n = + v e i n t e g e r w h e n \frac{n^{2} + n}{2} = r]$ $T_{r} = n + 1 [w h e n \frac{{(n + 1)}^{2} + n}{2} > r > \frac{n^{2} + n}{2}]$ $e x a m p l e f i n d T_{19} a n d T_{100} u s i n g a b o v e f o r m u l l a$ $s t e p . . . 1) p u t n = 1, 2, 3 . . . s o t h a t \frac{n^{2} + n}{2} a p p r o a c h e s t o 19$ $\frac{4^{2} + 4}{2} = 10$ $\frac{5^{2} + 5}{2} = 15 \leftarrow l o o k h e r e$ $\frac{6^{2} + 6}{2} = 21$ $n o w l o o k 19 > 15 w h e n n = 5$ $t h a t m e a n s$ $1, 2, 2, 3, 3, 3 . . . 5, 5, 5, 5, 5 \leftarrow t o t a l 15 t e r m s$ $1, 2, 2, 3, 3, 3 . . . . 5, 5, 5, 5, 5, 6, 6, 6, \underset{\underset{19 t h t e r m}{↑}}{6}, 6, 6 \leftarrow 21 t e r m s$ $p l s c h e c k . . .$
	Commented by gunawan last updated on 22/Feb/19

	$thank you Sir$ $I really appreciate$
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