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If(x 2 -x+1) -1 = a 0 +a 1 x +a 2 x 2 +a………………………,|x| m =0, if m is of the form: A) 3k B) 3k +1 c)3k+2 d)2k Please help.

If(x2-x+1)-1 = a0 +a1x +a2x2+a………………………,|x|<1, then am=0, if m is of the form:


A)     3k       B) 3k +1      c)3k+2      d)2k


 


Please help.

Grade:11

2 Answers

Akash Kumar Dutta
98 Points
8 years ago

Dear Prajwal,

These sums are not be solved in computer.but then also i am writing the solution.
Hope you understand...

expanding the equation we get = 1  - x(x-1) + [x(x-1)]^2+.... [(-1)x(x-1)]^m/2 +  [(-1)x(x-1)]^(m+2)/2 ......[(-1)x(x-1)]^m ... to infinite.
we do not need to go furthur since there are no more terms having the term x^m.
expand the terms internally and getting all the terms of x^m...

So taking common all the coefficients we get:-
 x^m . { (-1)^m/2 +  (-1)^(m+2)/2 . (m+2)/2C4 +.........(-1)^(m-1).  (m-1)C1 + (-1)^m }
Heance for the coefficient to be zero the subsequent terms should be of different signs.

hence the first and last terms,..second and second last .......should be of different signs.
taking the first and last...
they will be of different signs only if m is even and k is odd.
hence m is 2k.
its not 3k+1 since its not even always e.g. for k=2 , 3k+1=7(not even)

SO 2k IS THE ANSWER (d)

Regards.

Akash Kumar Dutta
98 Points
8 years ago

Dear Prajwal,

I AM VERY SORRY...AFTER DOING IT AGAIN I REALISED A MISTAKE...
I THINK THE ANSWER THAT I GAVE AT FIRST WAS WRONG...
SO HERES ANOTHER ATTEMT:-

(x^2 - x+1)^-1 = [ (x^3 + 1) / (x+1) ]^-1
                         = (x+1) . [(x^3 + 1)^-1]
                         = (x+1) . [ 1 -x^3 + x^6 - x^9 + x^12...... till infinity]
                         = { x - x^4 + x^7 -x^10.........+ 1-x^3 + x^6 - x^9......... }
                         = { 1 +x - x^3 - x^4 + x^6 + x^7 - x^9 - x^10................}
                         Hence all the terms are in the forms... 3k,3k+1,2k.
BUT NONE OF THEM ARE IN THE FORM 3k+2.

HENCE FOR a(m)=0,

m=3k+2 (c) (ANS)

HOPE THIS HELPS.

Regards.

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