InterviewSolution
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InterviewSolution
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The expressions `1+x,1+x+x^2,1+x+x^2+x^3.............1+x+x^2+..............+x^a` are mutiplied together and the terms of the product thus obtained are arranged in increasing powers of `x` in the from of `a_0+a_1x+a_2x^2+.................,` then sum of even coefficients?A. `20!`B. `21!`C. `(21!)/(2)`D. `19!` |
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Answer» Correct Answer - C `(c )` Let `f(x)=(1+x)(1+x+x^(2))(1+x+x^(2)+x^(3))…(1+x+x^(2)+….+x^(20))` Highest degree of `f(x)` `=` highest degree of `x` present in the `f(x)` `=1+2+3+…+20` `=(20(21))/(2)=210` Since in the expansion of `f(x)` degree of `x` from zero to `210`, all present so all terms containing `x^(0)`, `x^(1)`, `x^(2)`.....`x^(210)` will be present `:.` total no. of terms `=210+1=211` `(1+x)(1+x+x^(2))....(1+x+x^(2)+....+x^(20))` `=a_(0)+a_(1)x+a_(2)x^(2)+....+a_(210)x^(210)`......`(i)` Replacing `x` by `1//x`, we get `(1+(1)/(x))(1+(1)/(x)+(1)/(x^(2)))....(1+(1)/(x)+(1)/(x^(2))+....+(1)/(x^(20)))` `=a_(0)+(a_(1))/(x)+(a_(2))/(x^(2))+....+(a_(210))/(x_(210))` Taking `L.C.M` both sides, we get `(x+1)(x^(2)+x+1)....(x^(20)+x^(19)+....+x+1)` `=a_(0)x^(210)+a_(1)x^(209)+...+a_(209)x+a_(210)` Thus `a_(0)+x^(210)+a_(1)x^(209)+....+a_(209)x+a_(210)` Thus `a_(r )=a_(210-r)` `a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+.....` `=(1+x)(1+x+x^(2))...(1+x+x^(2)+...+x^(20))` Putting `x=1` in `(i)` `:. a_(0)+a_(1)+a_(2)+....=21!`......`(ii)` Again putting `x=-1` `a_(0)-a_(1)+a_(2)-a_(3)+a_(4)=0`.........`(iii)` Adding equation `(ii)` and `(iii)`, we have `2[a_(0)+a_(2)+a_(4)+....]=21!` `:.a_(0)+a_(2)+a_(4)+....=(21!)/(2)=` sum of even coefficients |
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