"So the coefiicient will be the same for all ?? a^47,a^46 and so on??""
@ rahul ... a^49  sum of all roots a^48  sum of (product of two roots at a time). a^47  sum of product of 3 roots at a time . .... (if sign should consider then alternative + and _ signs from a^50). for Eg : (xk1)(xk2) = 0 is equation in variable x and k1, k2 are roots ( as like a is var & 1,2 are roots in the above equation) then equation is x^2  (k1+k2)x+k1k2 =0 i.e, x^2 coeffient is 1 x^1 coeffient is  sum of roots x^0 coeffient is  product of roots (or sum of product of roots two at a time ).... (as equation power is 2 it ended thr.). To: "Sachin K" <sachin.sk@gmail.com>
Cc: ascent4cat@yahoogroups.com Date: Sunday, 21 March, 2010, 7:40 PM Thnx Sachin..... Got it.... So the coefiicient will be the same for all ?? a^47,a^46 and so on?? Rahul On Sun, Mar 21, 2010 at 12:30 PM, Sachin K <sachin.sk@gmail.com> wrote: Hey Rahul, I will detail out the procedure here for you.. Any expression like this is a polynomial in 'a'.. and the numbers within the bracket (e.g. 1, 2, 3, ...50) are the roots of this polynomial in 'a', and you can also write this polynomial in its expsnaded form as:
(a1)(a2)(a 3)......( a50) = (a^50)  (SUM_OF_ROOTS) *(a^49) + (SUM_OF_ALL_ PRODUCT_OF_ TWO_ROOTS) *(a^48)  ....... + (PRODUCT_OF_ ROOTS). So, You can see yourself that the coefficient of a^48 is nothing but the "SUM_OF_ALL_PRODUCT_ OF_TWO_ROOTS". Answer = (1.2 + 1.3 + 1.4 + ... + 1.50) + (2.3 + 2.4 + .... + 2.50) + (3.4 + ... +3.50) + ...+ (48.49 + 48.50) + (49.50). GOT IT !! Sachin Kumar ============ ========= ========= ========= ========= ========= ========= Hey guyz.. how to find out the Coefficient of a^48 in (a1)(a2)(a 3)......( a50)?? Plz explain the answer with the method.... Thanks Rahul.
