∑k=1nkxk=Sn\sum_{k=1}^{n} kx^k = S_n∑k=1nkxk=Sn とおくと Sn=x+2x2+3x3+⋯+(n−1)xn−1+nxnS_n = x + 2x^2 + 3x^3 + \cdots + (n-1)x^{n-1} + nx^nSn=x+2x2+3x3+⋯+(n−1)xn−1+nxn この両辺に xxx をかけて xSn=x2+2x3+3x4+⋯+(n−1)xn+nxn+1xS_n = x^2 + 2x^3 + 3x^4 + \cdots + (n-1)x^n + nx^{n+1}xSn=x2+2x3+3x4+⋯+(n−1)xn+nxn+1 Sn−xSn=(1−x)Sn=x+x2+x3+⋯+xn−nxn+1=x(1−xn)1−x−nxn+1S_n - xS_n = (1-x)S_n = x + x^2 + x^3 + \cdots + x^n - nx^{n+1} = \frac{x(1-x^n)}{1-x} - nx^{n+1}Sn−xSn=(1−x)Sn=x+x2+x3+⋯+xn−nxn+1=1−xx(1−xn)−nxn+1 Sn=x(1−xn)−nxn+1(1−x)(1−x)2=x{1−(n+1)xn+nxn+1}(1−x)2S_n = \frac{x(1-x^n) - nx^{n+1}(1-x)}{(1-x)^2} = \frac{x\{1-(n+1)x^n + nx^{n+1}\}}{(1-x)^2}Sn=(1−x)2x(1−xn)−nxn+1(1−x)=(1−x)2x{1−(n+1)xn+nxn+1}