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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

Pspl
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abababa
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  • Start with proving the equality for $n=1$ and post it on your answer! You have to show us some effort to solve the question yourself :) – Pspl Dec 06 '19 at 10:08
  • Please show us your original steps - we may then be able to show you where you went wrong – lioness99a Dec 06 '19 at 10:08
  • ok, thanks couple of moments I will edit – abababa Dec 06 '19 at 10:09
  • @shahar Hint: Try to start by verifying the situation when $n$ is a small number, and then assume that the assumption $P_n$ is true and try to show that if $P_n$ is true then $P_{n+1}$ is also true. – Kevin.S Dec 06 '19 at 10:13
  • In the inductive step you basically want to show that $\frac n3 (2n+1)(4n+1) + (2n+1)^2 + (2n+2)^2 = \frac {n+1}3 (2(n+1)+1)(4(n+1)+1)$ – Deepak Dec 06 '19 at 10:15
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    ok so I edited it for the steps I did, and to where I got confused – abababa Dec 06 '19 at 10:39
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    For $n=k+1$ you have a typo.... Inside the parenthesis.... – dmtri Dec 06 '19 at 10:50
  • @shahar Sometimes proving a more general case can be somewhat easier than a more specific one. With your question, such as shown in Sum of Consecutive Squares, you can try using induction to prove $\sum_{i=1}^{m} i^2 = \frac{m(m+1)(m+2)}{6}$. Then you can replace $m$ with $2n$ to get the specific result you're trying to prove. By doing this, you may find the algebra somewhat easier to deal with. – John Omielan Dec 06 '19 at 13:45

1 Answers1

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\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

Fareed Abi Farraj
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