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Prove: 4 + 10 + 16 + ... + (6n -2) = n(3n + 1)

So this is my working out but my final answer shows that they don't equal ...but it should right? I can't see where I've gone wrong.

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Arthur
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Amanda_C
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    Note the difference between $3(a+b) = 3a+3b$ and $3a+b$. the error is when you substitute for $n$ by $k+1$. – Km356 Aug 19 '19 at 05:13
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    The “next” step should be add up to $6(k+1)-2$, which is the summand after $6k-2$. The way you have it, the $n=2$ case would be $4 + (2+1)$, instead of $4+10$. – Arturo Magidin Aug 19 '19 at 05:19
  • $6n-2$ is a formula of $n$th term. You plugged $n=1$ and $n=k$. And what will it be if you plug $n=k+1$? ... Right, it is $6(k+1)-2$. – farruhota Aug 19 '19 at 05:57

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You're supposed to add $6(k+1)-2=6k+4$, viz. $3k^2+k+6k+4=3k^2+7k+4$. Similarly, it needs to turn out to be $(k+1)(3(k+1)+1)=(k+1)(3k+4)$, which indeed it does.

J.G.
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  • why is it 6(k + 1) -2 instead of k + 1? – Amanda_C Aug 19 '19 at 05:28
  • @Amanda_C If I asked you to prove $\sum_{k=1}^nf(k)=g(n)$ by induction for some arbitrary functions $f,,g$, your inductive step would note $\sum_{k=1}^{n+1}f(k)=\left(\sum_{k=1}^nf(k)\right)+f(n+1)=g(n)+f(n+1)$, which you'd need to show is $g(n+1)$. You certainly won't add $n+1$ to the left-hand side, unless $f(n+1)=n+1$ i.e. $f$ is the identity function. – J.G. Aug 19 '19 at 05:41
  • so like, if the question was prove 1 + 3 + 5 + .... + (2n - 1) = blah or whatever, would I use 2(k + 1) instead of k + 1 because there's a 2 number gap between each number given e.g 1 3 and 5? – Amanda_C Aug 19 '19 at 07:18
  • @Amanda_C You'd use $2(k+1)-1$. (The answer's $n^2$, by the way.) – J.G. Aug 19 '19 at 07:55