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Please solve the problem from binomial theorem.
1)   Find the coefficient of x^65 in the expansion of x(x+1)^2(x+2)^3(x+3)^4.........(x+10)^11
2).  Find the coefficient of x^(n^2+n-14)/2 in the expansion of (x-1)(x^2-2)(x^3 -3).........(x^n-n)
9 months 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
9 months ago
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