C(k,4)（0.95）^k-4前N项和有公式吗?PS：C(k,4)是排列组合K个取4个

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/17 04:34:48

C(k,4)=k(k-1)(k-2)(k-3)/4!
∑[k=4..n]C(k,4)x^(k-4)
=1/(24x^4)∑[k=4..n]k(k-1)(k-2)(k-3)x^k
=1/(24x^4)∑[k=0..n]k(k-1)(k-2)(k-3)x^k
构造f(x)=∑[k=0..n]x^k=(1-x^(n+1))/(1-x)
f'(x)=∑[k=0..n]kx^(k-1)=[(1-x^(n+1))/(1-x)]'
f''(x)=∑[k=0..n]k(k-1)x^(k-2)=[(1-x^(n+1))/(1-x)]''
f'''(x)=∑[k=0..n]k(k-1)(k-2)x^(k-3)=[(1-x^(n+1))/(1-x)]'''
f""(x)=∑[k=0..n]k(k-1)(k-2)(k-3)x^(k-4)=[(1-x^(n+1))/(1-x)]""
对(1-x^(n+1))/(1-x)求4次导数得到:
∑[k=0..n]k(k-1)(k-2)(k-3)x^k
=1/24*(24*x^n+2*x^(n-3)*n+20*x^n*n^3-6*x^(n+1)*n^3-4*x^(n-2)*n^4+6*x^(n-1)*n^4-4*x^n*n^4+x^(n+1)*n^4-x^(n-3)*n^2+4*x^(n-2)*n^2-12*x^(n-2)*n+6*x^(n-1)*n^2+36*x^(n-1)*n-20*x^n*n^2-20*x^n*n+11*x^(n+1)*n^2-6*x^(n+1)*n+12*x^(n-2)*n^3-24*x^(n-1)*n^3-2*x^(n-3)*n^3+x^(n-3)*n^4-24)/(x-1)^5
将x=0.95代入得到:
∑[k=4..n]C(k,4)0.95^(k-4)
=-163698.3551*.95^(n+1)*(1.00268750*n+.625e-5*(n+1)^4+.43750e-3*(n+1)^3+.257188e-1*(n+1)^2+20.55083750)+3200000.000

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