An equation `a_0+a_1x+a_2x^2+...............+a_99x^99+x^100=0` has roots `.^(99)C_0 ,.^(99)C_1,.^(99)C_2....^(99)C_99`. Find the value of `a_99`.

Answer

A

Explanation

`a_(0) + a_(1)x+a_(2)x+"...."+a_(99)x+x^(100) = 0` has root `.^(99)C_(0), .^(99)C_(1), .^(99)C_(2), "....", .^(99)C_(99)`.
or `a_(0) + a_(1)x+a_(2)x^(2) + "...."+a_(99)x^(99) + x^(100)`
`= (x-.^(99)C_(0))(x-.^(99)C_(1)) (x-.^(99)C_(2))"....."(x-.^(99)C_(99))`
Now, sum of root is
`.^(99)C_(0)+.^(99)C_(1)+.^(99)C_(2)+"...."+.^(99)C_(99)= - (a_(99))/("coefficient of" x^(100))`
or `a_(99) = - 2^(99)`
Also, sum of product of roots taken two at a time is
`(a_(99))/("coefficient of" x^(100))`
`:. underset(0 le i lt j le 99)(sumsum) .^(99)C_(i).^(99)C_(j) = ((underset(i=0)overset(99)sumunderset(j=0)overset(99)sum.^(99)C_(i).^(99)C_(j))- underset(i=0)overset(99)sum(.^(99)C_(i))^(2))/(2)`
`= ((underset(i=0)overset(99)sum.^(99)C_(i)2^(99))-underset(i=0)overset(99)sum(.^(99)C_(i))^(2))/(2)`
`= (2^(99)2^(99)-underset(i=0)overset(99)sum(.^(99)C_(i))^(2))/(2)`
`= (2^(198) - .^(198)C_(99))/(2)`
`(.^(99)C_(0))^(2)+ (.^(99)C_(1))^(2) + "......" + (.^(99)C_(99))^(2)`
`= (.^(99)C_(0) + .^(99)C_(1) + .^(99)C_(2) "......." + .^(99)C_(99))^(2) - 2 underset(0lei ltjle99)(sumsum).^(99)C_(i).^(99)C_(j)`
`= (-a_(99))^(2) - 2a_(98)`
`= a_(99)^(2) - 2a_(98)`