elementary prove thru induction - dumb stumbling

Question

i am trying to prove this statement for all $n \in \mathbb{N}$ with the help of induction:

$4 \sum_{k=1}^{n} (-1)^kk=(-1)^n(2n+1)-1$

base case: n=1

$4 \sum_{k=1}^{1} (-1)^11=-4=(-1)^1(2*1+1)-1$ .. OK

induction hypothesis: for all $n \in \mathbb{N}$ let be $4 \sum_{k=1}^{n} (-1)^kk=(-1)^n(2n+1)-1$

inductive step: $n \rightarrow n+1$

$4 \sum_{k=1}^{n+1} (-1)^kk=4 \sum_{k=1}^{n} (-1)^kk+(-1)^{n+1}(n+1)=(-1)^n(2n+1)-1+(-1)^{n+1}(n+1)=...help...=(-1)^{n+1}(2(n+1)+1)-1$

i need help for $..help..$

thanks a lot

Answer 1

Is : $$4 \sum_{k=1}^{n+1} (-1)^kk=4 \sum_{k=1}^{n} (-1)^kk+4(-1)^{n+1}(n+1)=(-1)^n(2n+1)-1-(-1)^n(4n+4)=(-1)^n(2n+1-4n-4)-1=(-1)^n(-2n-3)-1=(-1)^{n+1}(2n+3)-1=(-1)^{n+1}(2(n+1)+1)-1.$$

Answer 2

Hint $ $ Verify both sides are solutions of the recurrence $\rm\: f(n) - f(n\!-\!1) = (-1)^n 4n,\,\ f(1) =4,$ therefore they are equal for all $\rm\:n\:$ by a simple induction (which amounts to the uniqueness theorem for such recurrences -- see my many posts on telescopy).

Remark $\ $ For completeness I sketch the trivial inductive proof of said uniqueness theorem, sometimes called the fundamental theorem of difference calculus.

Theorem $ $ If $\rm\:f_1,f_2,g,h\:$ are functions on $\Bbb N,\,$ and $\rm\:f_1,f_2\:$ satisfy $\rm\:f(n\!+\!1) \color{#C00}= g(n) f(n) + h(n)\:$ for all $\rm\:n\ge 1,\:$ then $\rm\: f_1(1) = f_2(1)\:\Rightarrow\:f_1(n) = f_2(n)\:$ for all $\rm\:n \ge 1.$

Proof $\ $ Induct on $\rm\:n.\:$ The base case $\rm\:n=1\:$ is true by hypothesis, and the inductive step is

$$\begin{align} \rm \color{blue}{f_1(n)} = \color{#0A0}{f_2(n)}\ \Rightarrow\ f_1(n\!+\!1) &\rm \,\color{#C00} =\, g(n)\,\color{blue}{f_1(n)}+h(n)\\[.3em] &\,=\,\rm g(n)\, \color{#0A0}{f_2(n)}+h(n)\\[.3em] & \,\rm \color{#C00} = f_2(n\!+\!1)\end{align}\qquad\qquad\qquad$$