Polynomial
Uchiha Sasuke
28/03/2019 at 10:52-
Lê Anh Duy 24/07/2018 at 17:08
I think you should say: f(x) is divided by ..., because f(x) is divisible when it doesn't have any remainder.
Uchiha Sasuke
25/05/2018 at 08:52Uchiha Sasuke
25/05/2018 at 08:46Uchiha Sasuke
25/05/2018 at 08:39-
Uchiha Sasuke 26/05/2018 at 01:19
Thank you very much
-
P(x) = a0 + a1x + a2x2 + ... + a45x45
=> P(1) = a0 + a1 + a2 + ... + a45 = (2 + 1 + 2)15 = 515
=> P(-1) = a0 - a1 + a2 - ... - a45 = (2 - 1 - 2)15 = (-1)15
==> P(1) - P(-1) = 2a1 + 2a3 + ... + 2a45 = 515 + 1
P(1) + P(-1) = 2a0 + 2a2 + ... + 2a44 = 515 - 1
=> (a1 + a3 + ... + a45) - (a0 + a2 + ... + a44) = \(\dfrac{5^{15}+1-5^{15}+1}{2}=1\)
=> S = -a0 + a1 - a2 + .... - a44 + a45 = 1
P/s: Not S in the topic
Uchiha Sasuke
26/01/2018 at 07:48tranthuydung
09/03/2017 at 18:52-
John 11/03/2017 at 09:07
Let \(Q\left(x\right)=\left(x+1\right)P\left(x\right)-x\), (*)
since \(P\left(n\right)=\dfrac{n}{n+1}\) we infer:
\(Q\left(n\right)=\left(n+1\right)P\left(n\right)-n=0\) for \(n=0,1,..,2015\).
Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:
\(Q\left(x\right)=a\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\) (**)
In other hand, in (*) we set x = -1 then \(Q\left(-1\right)=1\). And replace to (**) we have:
\(1=Q\left(-1\right)=a\left(-1\right)\left(-2\right)...\left(-2016\right)\)
\(\Rightarrow a=\dfrac{1}{2016!}\)
Finally \(Q\left(x\right)=\dfrac{1}{2016!}\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)
\(\Rightarrow Q\left(2016\right)=\dfrac{1}{2016!}2016.2015...1=1\)
from (*) \(Q\left(2016\right)=\left(2016+1\right)P\left(2016\right)-2016\)
\(\Rightarrow\left(2016+1\right)P\left(2016\right)-2016=1\)
\(\Rightarrow P\left(2016\right)=\dfrac{2017}{2017}=1\)
tranthuydung selected this answer. -
FA KAKALOTS 28/01/2018 at 22:05
Let Q(x)=(x+1)P(x)−x
, (*)
since P(n)=nn+1
we infer:
Q(n)=(n+1)P(n)−n=0
for n=0,1,..,2015
.
Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:
Q(x)=a(x−0)(x−1)...(x−2015)
(**)
In other hand, in (*) we set x = -1 then Q(−1)=1
. And replace to (**) we have:
1=Q(−1)=a(−1)(−2)...(−2016)
⇒a=12016!
Finally Q(x)=12016!(x−0)(x−1)...(x−2015)
⇒Q(2016)=12016!2016.2015...1=1
from (*) Q(2016)=(2016+1)P(2016)−2016
⇒(2016+1)P(2016)−2016=1
⇒P(2016)=20172017=1