Question-7101

Question Number 7101 by Tawakalitu. last updated on 10/Aug/16

Commented by Tawakalitu. last updated on 10/Aug/16

W o r k i n g p l e a s e \dots \dots \dots \dots

Answered by Yozzii last updated on 10/Aug/16

u (x) = \frac{\underset{n = 1}{\sum^{x}} n^{2} (x - n - 1)}{\underset{n = 1}{\sum^{x}} n^{3}} (x \in N)

[\underset{n = 1}{\sum^{x}} n^{3} = \frac{x^{2} {(x + 1)}^{2}}{4}, \underset{n = 1}{\sum^{x}} n^{2} = \frac{x (x + 1) (2 x + 1)}{6}]

u (x) = \frac{\underset{n = 1}{\sum^{x}} {n^{2} (x - 1) - n^{3}}}{x^{2} {(x + 1)}^{2} / 4}

u (x) = \frac{(x - 1) \frac{x (x + 1) (2 x + 1)}{6} - \frac{x^{2} {(x + 1)}^{2}}{4}}{x^{2} {(x + 1)}^{2} / 4}

u (x) = \frac{4 (x - 1) (2 x + 1)}{6 x (x + 1)} - 1

u (x) = \frac{2 (2 x^{2} - x - 1)}{3 x^{2} + 3 x} - 1

u (x) = \frac{4 - 2 x^{- 1} - 2 x^{- 2}}{3 + 3 x^{- 1}} - 1

\lim_{x \to \infty} u (x) = \frac{1}{3} ? ?

Commented by Tawakalitu. last updated on 10/Aug/16

I d o n t k n o w t h e a n s w e r b u t i k n o w y o u s o l v e v e r y w e l l

Commented by Tawakalitu. last updated on 10/Aug/16

T h a n k y o u v e r y m u c h f o r y o u r s u p p o r t .

i w i l l w r i t e i t \dots .

Commented by Tawakalitu. last updated on 10/Aug/16

i n c a s e t h e r e i s a n u p d a t e, i w i l l s t i l l l e a r n m o r e, t h a n k s s i r

Commented by Tawakalitu. last updated on 11/Aug/16

T h e a n s w e r i s c o r r e c t \dots . i h a v e s e e n t h e s c o r e, T h a n k s s o m u c h

G o d b l e s s y o u .