Accedmychevron_right11thchevron_rightmathchevron_rightSequences And Serieschevron_rightExercise 6.8

Let ( r=1+\dfrac{1}{n} ). Series to (n) terms: [ S=\sum_{k=1}^{n} k, r^{k-1} ]

Using (\sum_{k=1}^{n}kx^{k-1}=\dfrac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2})

With (x=1+\frac1n), after simplification one obtains (S=n^2).

Proved.