Help proving using Mathematical Induction (novice)

Question

I need to prove expression using mathematical induction $P(1)$ and $P(k+1)$, that:

$$ 1^2 + 2^2 + \dots + n^2 = \frac{1}{6}n(n + 1)(2n + 1) $$

Proving $P(1)$ gave no difficulties, however I was stuck with $P(k+1)$, I've reached this point:

$$ 1^2 + \dots + (k+1)^2 = 1^2 + \dots + k^2 + (k+1)^2 = \\ \frac{1}{6}k(k+1)(2k+1) + (k+1)^2 $$

I've checked answer from the exercise book, the next step would be:

$$ = \frac{1}{6}(k+1)(k(2k+1)+6(k+1)) $$

How it was converted like that? Could you provide some explanation?

Thank you in advance

Answer 1

If you look carefully at $\frac16k(k+1)(2k+1)+(k+1)^2$, you'll see that it's a sum of two terms, $\frac16k(k+1)(2k+1)$ and $(k+1)^2$. Both of these terms have a factor $(k+1)$, so we can factor it out: $$=(k+1)\left(\frac16k(2k+1)+(k+1)\right)$$ Next, we can move the $\frac16$ as follows: $$=\frac16(k+1)\left(k(2k+1)+6(k+1)\right).$$ And there you have it!

In general, in these situations, the trick is to factor or develop expressions in the right way. If you can't figure out how to go from expression A to expression B, maybe going from B to A can be easier.

Answer 2

\begin{align} \frac16 k(k+1)(2k+1) + (k+1)^2 &= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} \\ &=\frac{(k+1)}{6} \cdot \left(k(2k+1) + 6(k+1) \right)\\ &= \frac{k+1}{6} \cdot (2k^2+7k+6)\\ &= \frac{k+1}{6} \cdot (2k+3)(k+2)\\ \end{align}

where we first make them have the same denominator then factor $k+1$ out.

Answer 3

True for $n=1.$

Let $n \geq 1$ and suppose true for $ n. $

Then

$1^2 + 2^2 + ...+ n^2 + (n+1)^2 = \frac{1}{6}n(n+1)(2n+1)+ (n+1)^2 = \frac{1}{6}(n+1)(2n^2 + n + 6n +6)= \frac{1}{6}(n+1)[(n+1)+1][2(n+1)+1]$

$ \Box $

Answer 4

$\frac{1}{6}k(k+1)(2k+1) + (k+1)^2=$

$\frac{1}{6}k(k+1)(2k+1) + \frac 16\cdot 6(k+1)^2=$

$\color{green}{\frac{1}{6}}k\color{blue}{(k+1)}(2k+1) + \color{green}{\frac 16}\cdot 6\color{blue}{(k+1)}^2=$

.....Now just factor out $\color{green}{\frac{1}{6}}\color{blue}{(k+1)}$ and you get....

$\color{green}{\frac{1}{6}}k\color{blue}{(k+1)}(2k+1) + \color{green}{\frac 16}\cdot 6\color{blue}{(k+1)}^2=$

$\color{green}{\frac{1}{6}}\color{blue}{(k+1)}[k(2k+1)+6(k+1)]$

... Now from here we need to convert $[k(2k+1)+6(k+1)]\implies ((k+1)+1)(2(k+1)+1)=(k + 2)(2k+3)=2k^2 + 7k + 6$...

$=\frac 16(k+1)[k(2k+1)+6(k+1)]=$

$\frac 16(k+1)(2k^2 + 7k +6)=$

$\frac 16(k+1)[(2k^2 + 4k)+(3k+6)]=$

$\frac 16(k+1)[2k\color{red}{(k+2)} + 3\color{red}{(k+2)}]=$

$\frac 16(k+1)\color{red}{(k+2)}[2k+3]=$

$\frac 16(k+1)(k+2)(2k+3)=$

$\frac 16(\color{purple}{k+1})(\color{purple}{k+1} + 1)(2(\color{purple}{k+1})+1)$.

ANd we are done