f is a function verify f(x+1) +x^2 =3f(x) 1)find f(8) and f(12) 2) calculate Σ_(k=0) ^n f(k) 3) find Σ


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Question Number 48267 by maxmathsup by imad last updated on 21/Nov/18

$f i s a f u n c t i o n v e r i f y f (x + 1) + x^{2} = 3 f (x)$ $1) f i n d f (8) a n d f (12)$ $2) c a l c u l a t e \sum_{k = 0}^{n} f (k)$ $3) f i n d \sum_{k = 0}^{n} f^{2} (k) .$
Commented by maxmathsup by imad last updated on 22/Nov/18
$2) we have f(x+1)−f(x)=−x^2 +2f(x) ⇒ Σ_(k=0) ^n (f(k+1)−f(k))=−Σ_(k=0) ^n k^2 +2Σ_(k=0) ^n f(k) ⇒ f(n+1)−f(0) =−((n(n+1)(2n+1))/6) +2Σ_(k=0) ^n f(k) ⇒ 2Σ_(k=0) ^n f(k)=f(n+1)−f(0)+((n(n+1)(2n+1))/6) ⇒ Σ_(k=0) ^n f(k) =(1/2){ f(n+1)−f(0)+((n(n+1)(2n+1))/6)}$
$2) w e h a v e f (x + 1) - f (x) = - x^{2} + 2 f (x) \Rightarrow$ $\sum_{k = 0}^{n} (f (k + 1) - f (k)) = - \sum_{k = 0}^{n} k^{2} + 2 \sum_{k = 0}^{n} f (k) \Rightarrow$ $f (n + 1) - f (0) = - \frac{n (n + 1) (2 n + 1)}{6} + 2 \sum_{k = 0}^{n} f (k) \Rightarrow$ $2 \sum_{k = 0}^{n} f (k) = f (n + 1) - f (0) + \frac{n (n + 1) (2 n + 1)}{6} \Rightarrow$ $\sum_{k = 0}^{n} f (k) = \frac{1}{2} {f (n + 1) - f (0) + \frac{n (n + 1) (2 n + 1)}{6}}$
Commented by maxmathsup by imad last updated on 22/Nov/18

$1) l e t s u p p o s e f p o l y n o m i a l \Rightarrow f (x) = {a x}^{2} + b x + c \Rightarrow$ $a {(x + 1)}^{2} + b (x + 1) + c + x^{2} = 3 ({a x}^{2} + b x + c) \Rightarrow$ $a (x^{2} + 2 x + 1) + b x + b + c + x^{2} - 3 {a x}^{2} - 3 b x - 3 c = 0 \Rightarrow$ ${a x}^{2} + 2 a x + a - 2 b x + (1 - 3 a) x^{2} + b - 2 c = 0 \Rightarrow$ $(1 - 2 a) x^{2} + (2 a - 2 b) x + a + b - 2 c = 0 \Rightarrow$ $1 - 2 a = 0 a n d 2 a - 2 b = 0 a n d a + b - 2 c = 0 \Rightarrow$ $a = \frac{1}{2}, b = a, c = \frac{a + b}{2} \Rightarrow a = \frac{1}{2}, b = \frac{1}{2}, c = \frac{1}{2} \Rightarrow$ $f (x) = \frac{1}{2} x^{2} + \frac{1}{2} x + \frac{1}{2} \Rightarrow f (8) = \frac{8^{2}}{2} + \frac{8}{2} + \frac{1}{2} = 32 + 4 + \frac{1}{2} = 36 + \frac{1}{2} = \frac{73}{2}$ $f (12) = \frac{12^{2}}{2} + \frac{12}{2} + \frac{1}{2} = \frac{144}{2} + 6 + \frac{1}{2} = 78 + \frac{1}{2} = \frac{157}{2}$ $2) \sum_{k = 0}^{n} f (k) = \frac{1}{2} \sum_{k = 0}^{n} k^{2} + \frac{1}{2} \sum_{k = 0}^{n} k + \frac{1}{2} \sum_{k = 0}^{n} (1)$ $= \frac{1}{2} \frac{n (n + 1) (2 n + 1)}{6} + \frac{1}{2} \frac{n (n + 1)}{2} + \frac{1}{2} (n + 1)$ $= \frac{n (n + 1) (2 n + 1)}{12} + \frac{n (n + 1)}{4} + \frac{n + 1}{2} .$
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