Question

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

The sum is `=n^2`

Explanation:

This is an arithmetico-geometric series

`S=1+2(1+1/n)+3(1+1/n)^2+4(1+1/n)^3+..................(n-1)(1+1/n)^(n-2)+n(1+1/n)^(n-1)`............`(1)`

Multiply both sides by `(1+1/n)`

`(1+1/n)S=1(1+1/n)+2(1+1/n)^2+3(1+1/n)^3+..................(n-1)(1+1/n)^(n-1)+n(1+1/n)^(n)`....................`(2)`

Substract `(1)-(2)`

`S-(1+1/n)S=1+1(1+1/n)+1(1+1/n)^2+1(1+1/n)^3+..................(1)(1+1/n)^(n-1)-n(1+1/n)^(n)`

This a geometric series, of common ratio `(1+1/n)`, `-n(1+1/n)^(n)`

`S-S-S/n=-S/n=1+1(1+1/n)+1(1+1/n)^2+1(1+1/n)^3+..................(1)(1+1/n)^(n-1)-n(1+1/n)^(n)`

The sum of a geometric serie is

`S_n=a(1-r^n)/(1-r)`

`-S/n=1(1-(1+1/n)^n)/(1-(1+1/n))-n(1+1/n)^n`

`-S/n=1((1-(1+1/n)^n)/(1-1-1/n))-n(1+1/n)^n`

`-S/n=-n((1-(1+1/n)^n)-n(1+1/n)^n`

`-S/n=-n+n(1+1/n)^n-n(1+1/n)^n`

`S=n^2`

Answer 2

See below.

Explanation:

`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`

Answer 3

`n^2`

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`

Answer 4

Let `x=1+1/n`

Then the sum 1st n terms of the given series may abbreviated as follows

`S=sum_(k=1)^(k=n)kx^(k-1)`

`=>S=d/(dx)sum_(k=1)^(k=n)x^k`

`=>S=d/(dx)[(x(x^n-1))/(x-1)]`

`=>S=[((n+1)x^n-1)/(x-1)-(x^(n+1)-x)/(x-1)^2]`

`=[(((n+1)x^n-1)(x-1)-(x^(n+1)-x))/(x-1)^2]`

`=[((n+1)x^(n+1)-(n+1)x^n-x+1-x^(n+1)+x)/(x-1)^2]`

`=[(nx^(n+1)-(n+1)x^n+1)/(x-1)^2]`

`=[(x^n(nx-n-1)+1)/(x-1)^2]`

`=[(x^n(n(1+1/n)-n-1)+1)/(1+1/n-1)^2]`

`=[(x^n(n+1-n-1)+1)/(1/n)^2]`

`=[(x^nxx0+1)/(1/n)^2]`

So `=>S=n^2`

Answer 5

See below.

Explanation:

Now considering

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

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

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