Proof by Induction simplification

Question

I am self studying proof by induction and tried some tasks, but im stuck with this one and i can't comprehend the solution.

1. Proof by induction that, for all positive integers n, $\sum_{i=1}^{n} (-1)^i i^2 = \frac{(-1)^n n(n+1)}{2} $

Proof:

when n=1

$\sum_{i=1}^{1} (-1)^1 1^2 = \frac{(-1)^1 1(1+1)}{2} $

$-1=-1$

$\therefore$ true when $n=1$

Assume true when $n=k$

$\sum_{i=1}^{k} (-1)^i i^2 = \frac{(-1)^k k(k+1)}{2} $

when $n=k+1$

$\sum_{i=1}^{k+1} (-1)^i i^2 = \frac{(-1)^{k+1} (k+1)((k+1)+1)}{2} $

$\sum_{i=1}^{k} (-1)^i i^2+(-1)^{k+1} (k+1)^2 = \frac{(-1)^{k+1} (k+1)((k+1)+1)}{2} $

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

This is how far I got and I cant get get further with the left side. I think I'm missing some easy algebra. The solution's next step (for the left side of the eq.) was:

$= \frac{(-1)^k(k+1)}{2}\cdot(k-2(k+1))$

$= \frac{(-1)^k(k+1)}{2}\cdot(-k-2)$

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

Answer 1

\begin{align} \frac{(-1)^k k(k+1)}{2}+(-1)^{k+1} (k+1)^2 &= (-1)^k(k+1) \left( \frac{k}{2} + (-1)(k+1) \right)\\ &= (-1)^k(k+1) \left( \frac{k}{2} -k -1 \right)\\ &= (-1)^k(k+1) \left( -\frac{k}{2} -1 \right)\\ &= -(-1)^k(k+1) \left( \frac{k}{2} + 1 \right)\\ &= -(-1)^k(k+1) \frac{k+2}{2}\\ &= \frac{(-1)^{k+1}(k+1)((k+1)+1)}{2} \end{align}

Answer 2

So we can simplify this as a sum ( I'm doing this because I'm more comfortable with expended - its preference, not necessary to do this way)

$$\sum_{i=0}^n (-1)^ii^2 = (-1)^1\cdot1^2 + (-1)^2\cdot2^2 + \dots+ (-1)^n\cdot n^2 = \frac{(-1)^n n(n+1)}{2}$$

Let's look for the base case $n =1$

$$\text{P(1) : } (-1)^1 \cdot 1^2 = \frac{(-1)^1\cdot1(2)}{2}$$

$$-1 = -1$$, so statement holds for $n=1$

For $k$

$$(-1)^1\cdot1^2 + (-1)^2\cdot2^2 + \dots+ (-1)^k\cdot k^2 = \frac{(-1)^k k(k+1)}{2}$$

For $k+1$ we need to proof

$$(-1)^1\cdot1^2 + (-1)^2\cdot2^2 + \dots+ (-1)^{k+1}\cdot (k+1)^2 = \frac{(-1)^{(k+1)} (k+1)(k+2)}{2}$$

we know from the above statement for $k$ that, $$(-1)^1\cdot1^2 + (-1)^2\cdot2^2 + \dots+ (-1)^k\cdot k^2 = \frac{(-1)^k k(k+1)}{2}$$

now we will substitute it into the $k+1$ statement

so statement for $k+1$ can be written as follows

$$(-1)^1\cdot1^2 + (-1)^2\cdot2^2 + \dots+(-1)^k\cdot k^2+ (-1)^{k+1}\cdot (k+1)^2 = \frac{(-1)^{(k+1)} (k+1)(k+2)}{2}$$

following is what we need to proof now, show that LHS is equal to RHS, and proof is done.

$$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{(-1)^k k(k+1)}{2} + (-1)^{k+1}\cdot (k+1)^2 = \frac{(-1)^{(k+1)} (k+1)(k+2)}{2}}}$$ let's take the LHS of the statement and simplify it.

$$(-1)^k(k+1)\left\{\frac{k}{2} + (-1)(k+1)\right\}$$ $$(-1)^k(k+1)\left\{\frac{-k-2}{2}\right\}$$ $$\frac{(-1)^{(k+1)}(k+1)(k+2)}{2}$$