InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If (1+x+x^(2))^(20) = a_(0) + a_(1)x^(2) "……" + a_(40)x^(40), then following questions. The value of a_(0) + 3a_(1) + 5a_(2) + "……" + 81a_(40) is |
|
Answer» `161 xx 3^(20)` Replacing x by `1//x`, we get `(1+x1/x+1/(x^(2)))^(20) = underset(r=0)overset(40)suma_(r)(1/x)^(r )` or `(1+x+x^(2))^(20) = underset(r=0)overset(40)suma_(r)x^(40-r) ""(2)` SINCE (1) and (2) are same series, COEFFICIENT of `x^(r )` is (1) `=` coefficient of `x^(r)` in (2). `rArr a_(r) = a_(40-r)` In (1) Putting `x = 1`, we get `3^(20) = a_(0)+a_(1)+a_(2)+"...."+a_(40)` `= (a_(0)+a_(1)+a_(2)+"...."+a_(19))+a_(20)+(a_(21)+a_(n+2)+"..."+a_(40))` `= 2(a_(0)+a_(1)+a_(2)+"...."+a_(19))+a_(20)""( :' a_(r) = a_(40-r))` or `a_(0) + a_(1) + a_(2) + "......."+ a_(19) = 1/2 (3^(20)-a_(20)) = 1/2(9^(10) - a_(20))` Also, `a_(0)+3a_(1)+5a_(2)+81a_(40)` `= (a_(0)+81a_(40))+(3a_(1)+79a_(39))+"...."+(39a_(19)+43a_(21))+41a_(20)` `= 82(a_(0) + a_(1) + a_(2) + "......" + a_(19)) + 41a_(20)` `= 41 xx 3^(20)` `a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - a_(3)^(2) + "....."` suggests that we have to MULTIPLY the TWO expansions. Replacing x by `-1//x` in (1), we get `(1-1/x+1/(x^(2)))^(20) = a_(0) - (a_(1))/(x)+(a_(2))/(x^(2))-"...."+(a_(40))/(a_(40))` `rArr (1-x+x^(2))^(20) = a_(0)x^(40) - a_(1)x^(39) + a_(2)x^(38) - "....."a_(40)""(3)` Clearly, `a_(0)^(3) - a_(1)^(2) + a_(2)^(2) + "....."+ a_(0)^(2)` is the coefficeint of `x^(40)` in `(1+x+x^(2)) (1-x+x^(2))^(20)` = Coefficient of `x^(40)` in `(1+x^(2)+x^(4))^(20)` In `(1+x^(2)+x^(4))^(20)` replace `x^(2)`, by y, then the coefficientof `y^(20)` in `(1+y+y^(2))^(20)` is `a_(20)`. Hence `a_(0)^(2) - a_(1)^(2) -"......"+a_(40)^(2) = a_(20)` or `(a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - "....." - a_(19)^(2)) + a_(20)^(2) + (-a_(21)^(2) + "....." + a_(40)^(2)) = a_(20)` or `2(a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - "....." - a_(19)^(2)) + a_(20)^(2) = a_(20)` or `a_(0)^(2) - a_(1)^(2) -"......" - a_(19)^(2) = (a_(20))/(2)[1-a_(20)]` |
|