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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jan 2008 13:15:40 IST
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Find the coefficient of x201 in (x-1)(x2-2)(x3-3)(x4-4)......(x20-20)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Jan 2008 19:49:24 IST
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Degree of polynomial = 1+2+3+.....20 = 210
x9 should be dropped from x210 so as to get x201.
Different combinations in which we can get x9 is :
x9 = x2x3x4 x9 = x1x8 x9 = x2x7 x9 = x3x6 x9 = x4x5 and x9 itself
from the 1st one x9 = x2x3x4 :
if we multiply variable terms each from (x-1),(x5-5),(x6-6)...... and constants from (x2-2),(x3-3),(x4-4) we get coefficient of x210 for 1st combination which comes out to be (-2)(-3)(-4) = -24
similarly for 2nd combination : x9 = x1x8
coefficient is (-1)(-8)
you can evaluate for each combination.
net coefficient comes out to be (-2)(-3)(-4)+(-8)(-1)+(-7)(-2)+(-6)(-3)+(-4)(-5)+(-9)
= 27
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jan 2008 00:27:12 IST
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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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Sorry for typing mistakes, please try to understand the symbols ...
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Sprinkle |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jan 2008 18:28:07 IST
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how can coefficient of x201 become zero
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