摘要:x<0的解集为
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设f(x)=(x+1)n(其中n∈N+).
(1)若f(x)=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+…+an(x-1)n,求a0及Sn=a1+a2+a3+…+an;
(2)当n=2013,计算:
-2
+…+k
(-1)k-1+…+2013
(-1)2012.
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(1)若f(x)=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+…+an(x-1)n,求a0及Sn=a1+a2+a3+…+an;
(2)当n=2013,计算:
| C | 1 2013 |
| C | 2 2013 |
| C | k 2013 |
| C | 2013 2013 |