Question

20．设函数f（x）=x（1+x）ⁿ，则${C}_{n}^{0}$+2${C}_{n}^{1}$+3${C}_{n}^{2}$+4${C}_{n}^{3}$+…+n${C}_{n}^{n-1}$+（n+1）${C}_{n}^{n}$=（n+2）•2^n-1．

Answer 1

分析 由${kC}_{n}^{k}={nC}_{n-1}^{k-1}$，得${C}_{n}^{0}$+2${C}_{n}^{1}$+3${C}_{n}^{2}$+4${C}_{n}^{3}$+…+n${C}_{n}^{n-1}$+（n+1）${C}_{n}^{n}$=${C}_{n}^{0}{+C}_{n}^{1}+..{．C}_{n}^{n}$+（${C}_{n}^{1}+{2C}_{n}^{2}+{3C}_{n}^{3}+…{nC}_{n}^{n}$）=${C}_{n}^{0}{+C}_{n}^{1}+..{．C}_{n}^{n}$+n（${C}_{n-1}^{0}{+C}_{n-1}^{1}+..{．C}_{n-1}^{n-1}$）=2ⁿ+n•2^n-1即可

解答 解：∵${kC}_{n}^{k}={nC}_{n-1}^{k-1}$
∴${C}_{n}^{0}$+2${C}_{n}^{1}$+3${C}_{n}^{2}$+4${C}_{n}^{3}$+…+n${C}_{n}^{n-1}$+（n+1）${C}_{n}^{n}$=${C}_{n}^{0}{+C}_{n}^{1}+..{．C}_{n}^{n}$+（${C}_{n}^{1}+{2C}_{n}^{2}+{3C}_{n}^{3}+…{nC}_{n}^{n}$）
=${C}_{n}^{0}{+C}_{n}^{1}+..{．C}_{n}^{n}$+n（${C}_{n-1}^{0}{+C}_{n-1}^{1}+..{．C}_{n-1}^{n-1}$）=2ⁿ+n•2^n-1
=（n+2）•2^n-1
故答案为：（n+2）•2^n-1

点评 本题考查了${kC}_{n}^{k}={nC}_{n-1}^{k-1}$的应用，即二项式展开式系数之和的应用，属于中档题．