Induction problem with sum to n-1

Question

I am having trouble solving this induction problem because when I do the last step proving P(n+1), I do not know what exactly I need to substitute. Note that in the expression below k is a positive integer number. $\displaystyle\sum_{i=0}^{n-1} 2ki+n-1 = {(k+1)(n-1)(n)}$

Answer 1

Let , $$S_n=\sum_{i=1}^{n-1}(2ki+(n-1))$$ Now, $$S_1=0$$ And $$(k+1)(1-1)(1)=0$$ The formula is true for $n=1$,$$$$$$$$

Now , $$S_n=\color{red}{(n-1)}+\color{blue}{(2k+n-1)}+\color{green}{(6k+n-1)}+\cdots+(2k(n-1)+n-1)$$$$$$ $$S_{n+1}=\color{red}{(n)}+\color{blue}{(2k+n)}+\color{green}{(6k+n)}+\cdots+(2k(n)+n)$$$$$$ $$S_{n+1}-S_n=\color{red}{1}+\color{blue}{1}+\color{green}{1}+\cdots\text{n times}\cdots+1+(2kn+n)$$$$$$ $$S_{n+1}-S_n=n+(2kn+n)$$$$$$ $$\bbox[5px,border:2px solid red]{S_{n+1}-S_n=2(k+1)n}$$

If $S_n=(k+1)(n-1)n$ $$S_{n+1}=(k+1)n(n+1)$$ $$S_{n+1}-S_n=(k+1)(n)(n+1-n+1)$$ $$S_{n+1}-S_n=2(k+1)n$$ Which is also true,

Hence....