IIT JEE , AIEEE Forum - Mathematics , Algebra

sprinkle

hm.india!

given polynomial has a degree = 1+2+3+..........+20 = 210

we have to find out the coeff. of x^201

Now, when u get the value of the expression

(x-1)(x²-2)(x³-3)(x⁴-4)......(x²⁰-20)

you actually take either x (with some power) or the constant from every factor and keep multiplying till the end. u do like this until all possibilities are over. e.g. when u take constant from every factor, u get -1.-2.-3......-20 = 20! as one term, when u take x from first factor and constant from every factor u get - 19! x etc.

Now if u want x^201, u will take variable part (x with some power) from many factors and constant from a few ones. In this process, note that u can't leave x^20 in any case as then u get max power as 210 - 20 = 190, similarly u can't leave x^19, x^18 ....... x^10. max power x-term u can leave is x^9 because in that case if u take x-term from all factors, it will give x^201.

Thus u have to take x-term from all factors from (x^10 - 10) ..... onwards.

and u have to take some more x-terms and constant terms from the following so that it gives x^9:
(x-1)(x²-2)(x³-3)........(x^9-9)

Thus coeff. of x^201 in given expression (in the ques.)
= sum of all possible products of constants when corresponding x-terms make an x^9
= (-9) + (-8)(-1) + (-7)(-2) + (-6)(-3) + (-6)(-2)(-1) + (-5)(-4) + (-5)(-3)(-1) + (-4)(-3)(-2)

(please note the style of writing so that no term is missed)

= -9 + 8 + 14 + 18 -12 + 20 - 15 - 24 = 0

Thus ans is 0.

Sorry for any miscalculations!!! sorry for giving less time to group for last a few days. extremely busy now a days.....

iitkgp_bipin!

ur method is good, please try to give soln in clearer way and please avoid calculation mistakes. i think u left that case of -6.-2.-1 and -5.-3.-1....anyway......not a big deal..... good solution.....keep it up!

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