Alternating Coefficients

Algebra Level 4

Let P ( x ) = ( x 3 − x 2 + x + 4 ) 2001 = a 0 + a 1 x + a 2 x 2 + . . . . + a 6003 x 6003 P(x)=(x^{3}-x^{2}+x+4)^{2001}=a_{0}+a_{1}x+a_{2}x^{2}+....+a_{6003}x^{6003} .

The sum a 0 + a 2 + a 4 + . . . + a 6002 a_{0}+a_{2}+a_{4}+...+a_{6002} can be expressed in the form 5 b + c 2 \dfrac{5^{b}+c}{2} , where b and c are positive integers, and b is as large as possible. Find the value of 5 + b + c + 2 5+b+c+2 .


The answer is 2009.

View wiki
Sean Ty by Sean Ty , 21, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Raymond Lin
Jul 24, 2014

Note that P ( 1 ) = a 0 + a 1 + a 2 . . . . . + a 6003 P(1)=a_0+a_1+a_2.....+a_{6003} and P ( − 1 ) = a 0 − a 1 + a 2 . . . . . − a 6003 P(-1)=a_0-a_1+a_2.....-a_{6003} .

Then P ( 1 ) + P ( − 1 ) = 2 ( a 0 + a 2 + a 4 + . . . . . + a 6002 ) P(1)+P(-1)=2(a_0+a_2+a_4+.....+a_{6002}) , so a 0 + a 2 + a 4 + . . . . . + a 6002 = P ( 1 ) + P ( − 1 ) 2 = ( 1 − 1 + 1 + 4 ) 2001 + ( − 1 − 1 − 1 + 4 ) 2001 2 = 5 2001 + 1 2 a_0+a_2+a_4+.....+a_{6002}=\frac{P(1)+P(-1)}{2}=\frac{(1-1+1+4)^{2001}+(-1-1-1+4)^{2001}}{2}=\frac{5^{2001}+1}{2}

Therefore, b = 2001 b=2001 and c = 1 c=1 , so the answer is 2009 \fbox{2009} .

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Hemang Sarkar
Jul 22, 2014

put x=1,-1 and add.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Sean Ty As pointed out in the clarification, "P(x) has degree 6003, not 2001".

Can you check if you are receiving clarification emails? I've just sent another one as a test.

Calvin Lin Staff - 6 years, 10 months ago

Log in to reply

Oh! How careless of me. I'll edit it right away! Sorry for the misconceptions!

Were you meaning my Yahoo! Mails? Well I'm not really using it so I'm sorry if I didn't check on your emails. Sorry! I'll make sure to re-check my problems next time!

Sean Ty - 6 years, 10 months ago

Log in to reply

Do you want the sum to end at a 2000 a_{2000} ? I think you want it to end at a 6002 a_{6002} .

Also, a b + c d \frac{ a^b + c } { d} is somewhat ambiguous, since we could set a a to be many different values. Perhaps if you give the value of a a , and say that " b b is as large as possible"?

Yes, when other members request a clarification or a dispute, we send you an email stating what their comments were, which would allow you to easily troubleshoot.

Calvin Lin Staff - 6 years, 10 months ago

Log in to reply

@Calvin Lin – Right, sorry about that, I should have reviewed it more. Sorry for that!

Alright, I'll make sure to get on Yahoo! as soon as a new problem emerges.

Sean Ty - 6 years, 10 months ago

What about binomial theorem? Can it be used here?

Jayakumar Krishnan - 6 years, 10 months ago

Log in to reply

Way too hard

Figel Ilham - 6 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...