If-1-1-2-4-3-10-4-20-5-

Question Number 32987 by NECx last updated on 08/Apr/18

I f 1 = 1

2 = 4

3 = 10

4 = 20

5 = ?

Answered by MJS last updated on 09/Apr/18

one possibility :

a_{n} = \frac{1}{6} (n + 2) (n + 1) n

a_{5} = 35

with k given points a polynome

function of degree k - 1 is determined

in this case :

y = {a x}^{3} + {b x}^{2} + c x + d

1 = a + b + c + d

4 = 8 a + 4 b + 2 c + d

10 = 27 a + 9 b + 3 c + d

20 = 64 a + 16 b + 4 c + d

solving leads to

a = \frac{1}{6}; b = \frac{1}{2}; c = \frac{1}{3}; d = 0

Commented by MJS last updated on 09/Apr/18

\Rightarrow we can freely set a_{5} and find

a polynome of degree 4

Commented by MJS last updated on 09/Apr/18

if we set

y = e^{{a x}^{3} + {b x}^{2} + c x + d}

and solve as above we get

a_{5} = \frac{1024}{25}

y = \frac{a}{x^{3}} + \frac{b}{x^{2}} + \frac{c}{x} + d \Rightarrow a_{5} = \frac{3627}{125}

\dots

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