Aditya Gupta
Last Activity: 5 Years ago
Roots of x(x+1)^2...(x+10)^11 are
0, -1, -1, -2, -2, -2,..., -10, -10,...,-10
Also, the degree of the above polynomial is 1+2+...+11= 66. So coefficient of x^65 is the following sum:
∑r(r+1), r ranging from 0 to 10. This is because of viettas formulas taking the roots one at a time.
The above sum is easy to find coz ∑r^2 and ∑r already have standard formulas