IIT JEE , AIEEE Forum - Mathematics , Algebra - Prove that 1^2008 + 2^2008 + 3^2008 + ??.+2006^2008 + 2007^2008 is divisible by 2008

srinathmallela

Prove that 1²⁰⁰⁸ + 2²⁰⁰⁸ + 3²⁰⁰⁸ + …….+2006²⁰⁰⁸ + 2007²⁰⁰⁸ is divisible by 2008

THIS QUESTION WAS ASKED IN RAMAIAH 2007

celestine

tis is simple

use binomial thm to club the terms that add to 2008 as

X^2008 + 2008-X^2008 = 2008^2008 - other terms of 2008

akhil_o

aⁿ+bⁿ=(a+b)(a^n-1-a^n-2b+...b^n-1)

hence it is divisible by (a+b)

hence as celestine says divide into pairs of 2 such dat a+b=2008

each pair is divisible by 2008 hence the whole sum is divisble by 2008

RyuAmakusa

@celestine u will get 2 x^2008 what will u do to that.
@akhil_o i think that n should be odd for example 1^2+2^2 =5 5 is not divisible 3
plz..reply

hsbhatt

That formula is actually $x^n - y^n = (x-y) (x^{n-1}+x^{n-2}y+...+y^n)$ . (sorry: exp of y should be (n-1) )

Just check if the exponent is 2007 instead of 2008? Or ask Mr. Koteswar Rao!!

vineetnegi

me too

a^n+b^n

app remainder theorem

(a+b)=0

-a=b

a^n+(-a)^n

=0

if n is odd

else

2*a^N

vineetnegi

ya it should be an odd one
for
(2008-n)^2007-n^2007=sum of terms divisible by 2008

hsbhatt

which only leaves you with the task of proving that $1004^{2007}$ is divisible by 2008.

celestine

Ryuamakusa

I think u dint get my approach of doin it

all terms like 1,2007 2,2006 ......... pair up to give multiples of 2008
only 1004^2008 is left which is obviously divisible by 2008

celestine

What did u mean by 2X^2008

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