f-x-y-f-x-f-y-x-y-and-f-4-10-faind-f-2022-

Question Number 179230 by mathlove last updated on 26/Oct/22

f (x + y) = f (x) + f (y) + x \cdot y

a n d f (4) = 10 f a i n d f (2022) = ?

Answered by mr W last updated on 26/Oct/22

l e t x = 1, y = n - 1

f (n) = f (1) + f (n - 1) + n - 1

f (n - 1) = f (1) + f (n - 2) + n - 2

\dots . .

f (3) = f (1) + f (2) + 2

f (2) = f (1) + f (1) + 1

Σ :

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

f (4) = 4 f (1) + \frac{4 \times 3}{2} = 10

\Rightarrow f (1) = 1

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

f (2022) = \frac{2022 \times (2022 + 1)}{2} = 2045253

Commented by Acem last updated on 27/Oct/22

1 s t : W h y d i d y o u d o t h e s u m ? a s l o n g a s f (4) = 10

{d o e s n}^{'} t g i v e u s t h a t i t i s l i k e a n o t h e r f u n c t i o n

w h i c h c o m p t s t h e d e s c e n d i n g s u m o f i t v a r i a b l e .

W e {c a n}^{'} t k n o w t h a t u n l e s s w e c a l c u l a t e s o m e

v a l u e s t h e n w e c a n n o t i c e t h a t

f (x + y) = g (x, y) = \frac{(x + y) (x + y + 1)}{2}

2 n d : 1 + 2 + 3 + \dots + (n - 1) = \frac{n (n - 1)}{2} s o {w h a t}^{'} s

a b o u t s u m o f f (1) + f (2) + \dots f (n - 1)

3 r d : T h e s u m y o u m a d e {d o s e n}^{'} t r e p r e s e n t s t h e

m a i n f u n c t i o n . T h e f (4) + f (5) + \dots f (n)

i s s o m e t h i n g e l s e \dots

W e h a v e f (x + y) \dots . l i k e f (16), f (36) \dots e t c

Commented by DvMc last updated on 27/Oct/22

F o r y o u r s e c o n d p o i n t \dots I b e l i e v e t h a t

i n t h e r e s p o n s e o n l y w a s r e p l a c i n g

t h e v a l u e o f f (n - i) i n f (n - i + 1) \dots

Commented by mathlove last updated on 27/Oct/22

t h a n k s

Commented by mr W last updated on 27/Oct/22

i t r a n s f e r e d t h e q u e s t i o n t o f o l l o w i n g

r e c u r s i v e s e q u e n c e :

a_{n} = a_{1} + a_{n - 1} + n - 1

s o w e h a v e

a_{n - 1} = a_{1} + a_{n - 2} + n - 2

a_{n - 2} = a_{1} + a_{n - 3} + n - 3

\dots .

a_{3} = a_{1} + a_{2} + 2

a_{2} = a_{1} + a_{1} + 1

s u m a l l :

a_{n} + a_{n - 1} + a_{n - 2} + \dots + a_{3} + a_{2} = {n a}_{1} + a_{n - 1} + a_{n - 2} + \dots + a_{3} + a_{2} + \frac{n (n - 1)}{2}

\Rightarrow a_{n} = {n a}_{1} + \frac{n (n - 1)}{2}

Commented by Tawa11 last updated on 28/Oct/22

Great sir

Answered by FelipeLz last updated on 26/Oct/22

f (x + y) = f (x) + f (y) + x y

x = y = a \to f (2 a) = 2 f (a) + a^{2}

f (4) = 10

2 f (2) + 4 = 10

f (2) = 3

f (10) = f (8) + f (2) + 16 = 2 f (4) + f (2) + 32 = 55

f (20) = 2 f (10) + 100 = 210

f (50) = f (40) + f (10) + 400 = 2 f (20) + f (10) + 800 = 1275

f (100) = 2 f (50) + 2500 = 5050

f (200) = 2 f (100) + 10000 = 20100

f (500) = f (400) + f (100) + 40000 = 2 f (200) + f (100) + 80000 = 125250

f (1000) = 2 f (500) + 250000 = 500500

f (2000) = 2 f (1000) + 1000000 = 2001000

f (2022) = f (2000) + f (22) + 44000

f (2022) = 2001000 + f (20) + f (2) + 40 + 44000

f (2022) = 2001000 + 210 + 3 + 40 + 44000

f (2022) = 2045253

Commented by mathlove last updated on 27/Oct/22

t h a n k s f o r a l l

Answered by Acem last updated on 27/Oct/22

f (4) = f (2) + f (2) + 4 = 10 \Rightarrow f (2) = 3

f (2) = f (1) + f (1) + 1 = 3 \Rightarrow f (1) = 1

f (3) = f (1) + f (2) + 2 = 1 + 3 + 2 = 6 d o n c f (3) = 6

f (4) = 10

W e n o t i c e t h a t t h e f u n c t i o n f (x + y) r e p r e s e n t s

t h e d e s c e n d i n g s u m o f t h e s u m o f i t t w o v a r i a b l e s i . e .

x + y = 4 \to 4 + 3 + 2 + 1 = 10

x + y = 3 \to 3 + 2 + 1 = 6 \dots . e t c

i . e . f (x + y) = g (x, y) = \frac{1}{2} (x + y) (x + y + 1)

A n d f (2022) = 2 045 253

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