hi I started learning mathematical induction and I have a problem with a specific one
I need to prove that the left side is equal to the right side and I got lost on the way, so any answers will really help
$$ 1^2 + 2^2 + 3^2 + 4^2+….(2n)^2=\frac{n}{3}(2n+1)(4n+1)$$
edit:
this is my progress so far: $n=1$:
$$ 1^2 +2^2 =5=\frac{1}{3}\times3\times5$$
assumption $n=k$:
$$ 1^2 + 2^2 + 3^2 + 4^2+….(2k)^2=\frac{k}{3}(2k+1)(4k+1)$$
$n=k+1$
$$ 1^2 + 2^2 + 3^2 + 4^2+….(2k+1)^2=\frac{k+1}{3}(2k+3)(4k+5)$$
what I tried to do is opening the left side to look like the right side
$$ \frac{k}{3}(2k+1)(4k+1)+ (2k+1)^2 (2k+2)^2 =\frac{k}{3}8k^2+6k+1+8k^2+12k+5$$
and right from here is when I get lost, tried multiplying all the equation with 3
but really didnt get close
so thats what I tried to do any tips will help, and maybe you could try to tell me how should I try to look at both sides for future induction
Best Answer
\begin{align}1^2 + 2^2 + 3^2 + 4^2+\cdots+ [2(k+1)]^2&=1^2 + 2^2 + 3^2 + 4^2+\cdots(2k)^2+(2k+1)^2+(2k+2)^2 \\&=1^2 + 2^2 + 3^2 + 4^2+\cdots(2k)^2+4k^2+4k+1+4k^2+8k+4\\&=\frac{k}{3}(2k+1)(4k+1) +8k^2+12k+5\end{align}
On the other hand,
$\begin{align} \frac{k+1}{3}(2k+3)(4k+5) &= \frac{k}{3}(2k+3)(4k+5) +\frac{1}{3}(2k+3)(4k+5) \\ &=\frac{k}{3}(2k+1)(4k+5) +\frac{k}{3}(2)(4k+5)+\frac{1}{3}(2k+3)(4k+5) \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{k}{3}(2k+1)(4)+ \frac{k}{3}(2)(4k+5) + \frac{1}{3}(2k+3)(4k+5) \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[k(2k+1)(4)+ k(2)(4k+5) +(2k+3)(4k+5)] \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[4k(2k+1)+ 4k(2k+1)+6k +(2k+3)(4k+5)] \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[8k(2k+1)+6k +(2k+1)(4k+5)+2(4k+5)] \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[(12k+5)(2k+1)+6k+8k+10]\\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[(12k+5)(2k+1)+14k+10] \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[5(2k+1)^2+4k^2+16k+10] \\ &=\frac{k}{3}(2k+1)(4k+1) +\frac{1}{3}[24k^2+36k+15] \\ &=\frac{k}{3}(2k+1)(4k+1) +8k^2+12k+5 \\ &=1^2 + 2^2 + 3^2 + 4^2+\cdots+ [2(k+1)]^2 \end{align}$
And we're done