Saved Bookmarks
| 1. |
If n is a positive integer and (1 + x + x^(2))^(n) = a_(0) + a_(1) x + …. + a_(2 n) x^(2 n). Then, show that, a_(0)^(2) - a_(1)^(2) + ….a_(2n)^(2) = a_(n). |
|
Answer» Solution :`(1 + x + x^(2))^(N) = a_(0) + a_(1)x + ...+ a_(2N)x^(2n)…(i)` REPLACING x by `-1//x` , we get `(1- (1)/( x )+(1)/ x^(2))^(n) = a_(0) - (a_(1))/(x)+(a_(2))/(x^(2))+(a_(3))/(x^(3)) + ...+ (a_(2n))/(x^(2n))….(ii)` Now , `a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - a_(3)^(2) +...a_(2n)^(2) = ` , COEFFICIENT of the term independent of x in ` [ a_(0) + a_(1)x + a_(2) x^(2) +...+ a_(2n)x^(2n)] xx[a_(0) - (a_(1))/(x) + (a_(2))/(x^(2))- ... + (a_(2n))/(x^(2n))]` = Coefficient of the term independent of x in `(1 + x + x^(2))^(n) (1 - (1)/(x) + (1)/(x^(2)))^(n)` Now , RHS = ` (1 + x + x^(2))^(n) (1 - (1)/(x) + (1)/(x^(2)))^(n)` `((1 + x + x^(2))^(n) (x^(2) - x + 1 )^(n))/(x^(2n))=([(x^(2) + 1)^(2)- x ^(2)]^(n))/(x^(2n))` `((1 + 2x^(2) + x^(4) - x ^(2))^(n))/(x^(2n)) = (1 + x^(2) +x^(4))^(n)/(x^(2n))` Thus , ` a_(0)^(2) - a_(1)^(2) + a_(2)^(2) - a_(3)^(2) +...+ a_(2n)^(2)` = Coefficient of the term independent of x in ` (1)/(x^(2n)) (1 + x^(2) + x^(4))^(n)` = Coefficient of `x^(2n) ` "in" `(1 + x ^(2) + x^(4))^(n)` = COFFICIENT of ` t^(n)` "in" ` (1 + t + t^(2))^(n) = a_(n)`. |
|