Show-that-1-2-2-2-3-2-n-n-1-2-1-2-2-2-2-3-n-2-n-1-3n-5-3n-1-

Question Number 45932 by Tawa1 last updated on 18/Oct/18

Show that : \frac{1 . 2^{2} + 2 . 3^{2} + \dots + n {(n + 1)}^{2}}{1^{2} . 2 + 2^{2} . 3 + \dots + n^{2} (n + 1)} = \frac{3 n + 5}{3 n + 1}

Commented by math khazana by abdo last updated on 19/Oct/18

S_{n} = \frac{\sum_{k = 1}^{n} k {(k + 1)}^{2}}{\sum_{k = 1}^{n} k^{2} (k + 1)} = \frac{\sum_{k = 1}^{n} k (k^{2} + 2 k + 1)}{\sum_{k = 1}^{n} k^{3} + \sum_{k}^{n} k^{2}}

= \frac{\sum_{k = 1}^{n} k^{3} + 2 \sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k}{\sum_{k = 1}^{n} k^{3} + \sum_{k = 1}^{n} k^{2}}

= \frac{\frac{n^{2} {(n + 1)}^{2}}{4} + 2 \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}}{\frac{n^{2} {(n + 1)}^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{6}}

= \frac{\frac{n (n + 1)}{2} + \frac{2}{3} (2 n + 1) + 1}{\frac{n (n + 1)}{2} + \frac{2 n + 1}{3}}

= \frac{3 n (n + 1) + 4 (2 n + 1) + 6}{3 n (n + 1) + 4 n + 2}

= \frac{3 n^{2} + 3 n + 8 n + 4 + 6}{3 n^{2} + 3 n + 4 n + 2} = \frac{3 n^{2} + 11 n + 10}{3 n^{2} + 7 n + 2}

r o o t s o f 3 n^{2} + 7 n + 2

Δ = 49 - 4.3.2 = 49 - 24 = 25 \Rightarrow n_{1} = \frac{- 7 + 5}{6} = - \frac{1}{3}

n_{2} = \frac{- 7 - 5}{6} = - 2

r o o t s o f 3 n^{2} + 11 n + 10 \to Δ = 121 - 4.3.10

= 121 - 120 = 1 \Rightarrow n_{3} = \frac{- 11 + 1}{6} = - \frac{5}{3}

n_{3} = \frac{- 11 - 1}{6} = - 2 \Rightarrow

S_{n} = \frac{3 (n + \frac{1}{3}) (n + 2)}{3 (n + \frac{5}{3}) (n + 2)} = \frac{3 n + 1}{3 n + 5} a n d t h e r e i s a e r r o r

a t t h e Q .

Commented by math khazana by abdo last updated on 19/Oct/18

e r r o r a t t h e f i n a l l i n e S_{n} = \frac{3 (n + \frac{5}{3}) (n + 2)}{3 (n + \frac{1}{3}) (n + 2)}

= \frac{3 n + 5}{3 n + 1} a n d t h e r i s n o e r r o r a t t h e Q \dots

Commented by Tawa1 last updated on 19/Oct/18

God bless you sir

Commented by maxmathsup by imad last updated on 19/Oct/18

y o u a r e w e l c o m e s i r

Answered by math1967 last updated on 19/Oct/18

S_{n} = 1 . 2^{2} + 2 . 3^{2} + \dots . . + n {(n + 1)}^{2}

= t_{1} + t_{2} \dots \dots . + n^{3} + 2 n^{2} + n

∴ t_{1} + t_{2} + \dots \dots t_{n} = (1^{3} + 2^{3} + \dots n^{3}) + 2 (1^{2} + 2^{2} + . . + n^{2}) +

(1 + 2 + . . n)

∴ S_{n} = \frac{n^{2} {(n + 1)}^{2}}{4} + 2 \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}

= \frac{n (n + 1) (3 n + 5) (n + 2)}{12}

A g a i n 1^{2} . 2 + 2^{2} . 3 + \dots n^{2} (n + 1)

∴ t_{n} = n^{3} + n^{2}

∴ S_{n} = (1^{3} + 2^{3} + \dots . + n^{3}) + (1^{2} + 2^{2} + . . + n^{2})

= \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{6}

= \frac{n (n + 1) (n + 2) (3 n + 1)}{12}

∴ \frac{1 . 2^{2} + 2 . 3^{2} + \dots . n {(n + 1)}^{2}}{1^{2} . 2 + 2^{2} . 3 + \dots . . n^{2} (n + 1)}

\frac{n (n + 1) (3 n + 5) (n + 2)}{12} \times \frac{12}{n (n + 1) (n + 2) (3 n + 1)}

= \frac{3 n + 5}{3 n + 1} (p r o v e d)

Commented by Tawa1 last updated on 19/Oct/18

God bless you sir

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