Find the coefficient of X3 in the expansion of (1 + x) + (1 + x) 2 +. + (1 + x) 10

From the third term to the tenth term, the coefficient with x ^ 3 in each term is
C(3,3),C(4,3)...C(10,3)
C(n,3)=n!/[(n-3)!3!]
=>Coefficient sum = 1 + 4 + 10 + 20 + 35 + 56 + 84 + 120 = 330
Another approach:
（1+x）+（1+x）2+.+(1+x）10=[(1+x)^11-1]/(1+x-1)=[(1+x)^11-1]/x
This problem is equivalent to finding the x ^ 4 coefficient of (1 + x) ^ 11 = 11! / (7! 4!) = 330

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