Where did I go wrong in finding the sum of $1+2k+3k^2+...+nk^{n-1}$ using Abel's formula

The version of Abel's formula I'm using is $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)B_1+(a_2-a_3)B_2+...+(a_{n-1}-a_n)B_{n-1}+a_nB_n$ where $B_1=b_1$, $B_2=b_1+b_2$, $B_3=b_1+b_2+b_3$...$B_n=b_1+b_2+b_3+...+b_n$ and $a_n$ and $b_n$ are two sets of real number sequences.

Letting $a_1=1, a_2=2, a_3=3$ etc. and $b_1=1, b_2=k, b_3=k^2$ etc. the sum becomes $(1-2)(1)+(2-3)(1+k)+(3-4)(1+k+k^2)+...+(n-1-n)(1+k+k^2+...+k^{n-1})+n(1+k+k^2+...+k^n)$

Moving the last term to the front, it's equal to $n\left(\frac{k^n-1}{k-1}\right)-(1+(1+k)+(1+k+k^2)+...+(1+k+k^2+...+k^{n-1}))$

$=n\left(\frac{k^n-1}{k-1}\right)-\left(1+\frac{k^2-1}{k-1}+\frac{k^3-1}{k-1}+...+\frac{k^n-1}{k-1}\right)$

$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(k-1+k^2-1+k^3-1+...+k^n-1)$

$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(k+k^2+k^3+...+k^n+n(-1))$

$=n\left(\frac{k^n-1}{k-1}\right)-\frac{1}{k-1}(\frac{k(k^n-1)}{k-1}-n)$

$=\frac{nk^n}{k-1}-\frac{k(k^n-1)}{(k-1)^2}$

However, I seem to have an extra multiple of $k$, the answer should be $=\frac{nk^n}{k-1}-\frac{(k^n-1)}{(k-1)^2}$ (I've doubled checked my result against $1+2(2)+3(2)^2+4(2)^3+5(2)^4$ and it doesn't get the correct answer) but I can't spot where I went wrong.

Solution 1:

Hint:

With the correct expression $b_n=k^{n-1}$ instead of $b_n=k^n$ you will obtain the desired result.

Observe also that you wrongly simplified the last term $a_nB_n$ in your expression (these two errors have eliminated each other to give correctly the first term in the final expression).