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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) 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} .