Question-199547

Question Number 199547 by hardmath last updated on 05/Nov/23

Answered by mr W last updated on 05/Nov/23

1)

n \times n! = (n + 1 - 1) \times n! = (n + 1)! - n!

1 \times 1! + 2 \times 2! + \dots + 2022 \times 2022!

= 2! - 1! + 3! - 2! + \dots . + 2023! - 2022!

= 2023! - 1!

\Rightarrow a b = 2023 \times 1 = 2023

2)

f (a + b) = f (a) + f (b) + a b

f (2 a) = 2 f (a) + a^{2}

f (2) = 2 f (1) + 1^{2} = 5 \Rightarrow f (1) = 2

f (4) = 2 f (2) + 2^{2} = 2 \times 5 + 4 = 14

f (5) = f (4) + f (1) + 4 \times 1 = 14 + 2 + 4 = 20

f (10) = 2 f (5) + 5^{2} = 2 \times 20 + 25 = 65

f (11) = f (10) + f (1) + 10 \times 1 = 65 + 2 + 10 = 77

o r g e n e r a l l y f (x) = \frac{x (x + 3)}{2}

\Rightarrow f (11) = \frac{11 \times 14}{2} = 77

Commented by hardmath last updated on 05/Nov/23

thank you dear professor cool

Commented by hardmath last updated on 05/Nov/23

t h a n k y o u d e a r p r o f e s s o r p e r f e c t s o l u t i o n s

Commented by hardmath last updated on 05/Nov/23

dear professor,

how did you get the generalized solution

and is it always true ?

Commented by mr W last updated on 05/Nov/23

s i n c e f (x + y) = f (x) + f (y) + x y, y o u c a n

a s s u m e f (x) = {a x}^{2} + b x + c w i t h f (2) = 5 .

t h e n y o u g e t a = \frac{1}{2}, b = \frac{3}{2}, c = 0 .

Answered by AST last updated on 05/Nov/23

f (4) = 10 + 4 = 14 \Rightarrow f (8) = 44

5 = 2 f (1) + 1 \Rightarrow f (1) = 2; f (3) = 7 + 2 = 9

f (11) = 44 + 9 + 24 = 77

Commented by hardmath last updated on 05/Nov/23

t h a n k y o u p r o f e s s o r

Answered by ajfour last updated on 05/Nov/23

2 . f (x + h) - f (x) = f (h) + h x

f^{'} (x) = \frac{f (0)}{h} + x

\Rightarrow f (0) = 0

f (x) = \frac{x^{2}}{2} + k x

a s f (2) = \frac{2^{2}}{2} + 2 k = 5 \Rightarrow k = \frac{3}{2}

f (x) = \frac{x^{2}}{2} + \frac{3 x}{2} = \frac{x (x + 3)}{2}

Commented by hardmath last updated on 05/Nov/23

t h a n k y o u p r o f e s s o r

Commented by AST last updated on 05/Nov/23

f^{^{'}} (x) = x + c \Rightarrow f (x) = \int (x + c) = \frac{x^{2}}{2} + c x + k

f (0) = 0 \Rightarrow k = 0; f (2) = 5 \Rightarrow c = \frac{3}{2} \Rightarrow f (x) = \frac{x^{2} + 3 x}{2}

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