f-x-f-x-1-x-f-0-123-f-29-

Question Number 87352 by jagoll last updated on 04/Apr/20

f (x) = f (x + 1) - x

f (0) = 123

f (29) = ?

Commented by jagoll last updated on 04/Apr/20

x = 0 \Rightarrow 123 = f (1) - 0

f (1) = 123

x = 1 \Rightarrow 123 = f (2) - 1 \Rightarrow f (2) = 124

x = 2 \Rightarrow 124 = f (3) - 2 \Rightarrow f (3) = 126

x = 3 \Rightarrow 126 = f (4) - 3 \Rightarrow f (4) = 129

it correct

Commented by jagoll last updated on 04/Apr/20

f (29) = 123 + \frac{29 \times 28}{2}

= 123 + 406 = 529

Commented by jagoll last updated on 04/Apr/20

mr W, my work it correct ?

Commented by mr W last updated on 04/Apr/20

y e s

Commented by john santu last updated on 04/Apr/20

let f (x) = {ax}^{2} + bx + c

f (x + 1) = a {(x + 1)}^{2} + b (x + 1) + c

f (x + 1) - f (x) = x

2 ax + a + b = x

2 a = 1 \Rightarrow a = \frac{1}{2}

a + b = 0 \Rightarrow b = - \frac{1}{2}

f (x) = \frac{1}{2} x^{2} - \frac{1}{2} x + c

f (0) = c = 123

∴ f (x) = \frac{1}{2} x^{2} - \frac{1}{2} x + 123

then f (29) = \frac{1}{2} \times 29 (29 - 1) + 123

\frac{29 \times 28}{2} + 123

Commented by jagoll last updated on 04/Apr/20

thank you sir

Answered by mr W last updated on 04/Apr/20

f (k + 1) = f (k) + k

\underset{k = 0}{\sum^{n}} f (k + 1) = \underset{k = 0}{\sum^{n}} f (k) + \underset{k = 0}{\sum^{n}} k

f (n + 1) - f (0) + \underset{k = 0}{\sum^{n}} f (k) = \underset{k = 0}{\sum^{n}} f (k) + \underset{k = 0}{\sum^{n}} k

f (n + 1) - f (0) = \underset{k = 0}{\sum^{n}} k = \frac{n (n + 1)}{2}

f (n + 1) = f (0) + \frac{n (n + 1)}{2}

\Rightarrow f (n) = f (0) + \frac{n (n - 1)}{2}

\Rightarrow f (29) = 123 + \frac{29 \times 28}{2} = 529

Commented by jagoll last updated on 04/Apr/20

thank you sir

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