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........#?
5 Answers
The sum is
Explanation:
This is an arithmetico-geometric series
Multiply both sides by
Substract
This a geometric series, of common ratio
The sum of a geometric serie is
See below.
Explanation:
NOTE:
For
Explanation:
Here's a solution that uses no standard formula or differentiation:
Given:
#1+2*(1+1/n)+3*(1+1/n)^2+4*(1+1/n)^3+...+n(1+1/n)^(n-1)#
#= sum_(k=1)^n k(1+1/n)^(k-1)#
#= sum_(k=1)^n kx^(k-1)" "# where#x = 1+1/n = (n+1)/n#
Note that:
#(1-x) sum_(k=1)^(n-1) x^k#
#=sum_(k=1)^(n-1) x^k - sum_(k=2)^n x^k#
#=x+color(red)(cancel(color(black)(sum_(k=2)^(n-1) x^k))) - color(red)(cancel(color(black)(sum_(k=2)^(n-1) x^k))) - x^n#
#=x-x^n#
So:
#sum_(k=1)^(n-1) x^k = (x-x^n)/(1-x)#
We find:
#(1-x)sum_(k=1)^n kx^(k-1)#
#=sum_(k=1)^n kx^(k-1) - x sum_(k=1)^n kx^(k-1)#
#=sum_(k=1)^n kx^(k-1) - sum_(k=2)^(n+1) (k-1)x^(k-1)#
#=1+sum_(k=2)^n kx^(k-1) - sum_(k=2)^n (k-1)x^(k-1) - nx^n#
#=1-nx^n+sum_(k=2)^n x^(k-1)#
#=1-nx^n+sum_(k=1)^(n-1) x^k#
#=1-nx^n+(x-x^n)/(1-x)#
#=((1-nx^n)(1-x)+x-x^n)/(1-x)#
#=(1-color(red)(cancel(color(black)(x)))-nx^n+nx^(n+1)+color(red)(cancel(color(black)(x)))-x^n)/(1-x)#
#=(1-(n+1)x^n+nx^(n+1))/(1-x)#
So:
#sum_(k=1)^n kx^(k-1)= (1-(n+1)x^n+nx^(n+1))/(1-x)^2#
#color(white)(sum_(k=1)^n kx^(k-1)) = (1-(n+1)((n+1)/n)^n+n((n+1)/n)^(n+1))/(1-(1+1/n))^2#
#color(white)(sum_(k=1)^n kx^(k-1)) = n^2(1-(n+1)((n+1)/n)^n+n((n+1)/n)^(n+1))#
#color(white)(sum_(k=1)^n kx^(k-1)) = n^2(1-color(red)(cancel(color(black)((n+1)^(n+1)/n^n)))+color(red)(cancel(color(black)((n+1)^(n+1)/n^n))))#
#color(white)(sum_(k=1)^n kx^(k-1)) = n^2#
Let
Then the sum 1st n terms of the given series may abbreviated as follows
So
See below.
Explanation:
Now considering