## University Calculus: Early Transcendentals (3rd Edition)

$\Sigma_{n=0}^\infty (n+1)(n+2)x^{n}$
Since, we know that the Taylor Series for $\dfrac{1}{(1-x)^2}$ is defined as: $\dfrac{1}{(1-x)^2}=1+2x+3x^2+.....nx^{n-1}$ Now, $(\dfrac{1}{(1-x)^2})'=[1+2x+3x^2+.....nx^{n-1}]'=2+6x+12x^2+....n(n+1)x^{n-1}$ $=\Sigma_{n=0}^\infty (n+1)(n+2)x^{n}$