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Find the sum of the first n terms of the series : #1 + 2(1+1/n) + 3(1+1/n)^2 + 4(1+1/n)^3.......#?

Answer 1

Ariana Alvarez

See below.

#sum_(k=0)^n (k+1)x^k = d/(dx) sum_(k=1)^(n+1) x^k = #

#= d/(dx)((x^(n+2)-1)/(x-1)-1)=(x^(n+1) (n (x-1) + x-2)+1)/(x-1)^2# and making #x = 1+1/n# we get at

#sum_(k=0)^n (k+1)(1+1/n)^k =n^2 + (1 + 1/n)^n (1 + n)#

NOTE:

For #n=3#

#1 + 2 (1 + 1/3) + 3 (1 + 1/3)^2 + 4 (1 + 1/3)^3=499/27#

#3^2 + (1 + 1/3)^3 (1 + 3)=499/27#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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