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suppose we want to prove by induction that
$$1^2+ 2^2+ 3^2+ 4^2+ ....... + n^2= \frac{n}{6}(n+1)(2n+1)$$ it is very easy where we assume $k$ and $k+1$.

suppose we want to prove $$3^k>2^k, \forall k\in\mathbb R^+$$
then while doing induction (if we are allowed to), can we assume for any general $k$ and instead of proving it for $k+1$, are we allowed to do it for limit as $h$ tends to zero $k+h$ ?

PNT
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Dusty_Wanderer
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  • This is not clear. You generally can't do induction in $\mathbb R$. Is that what you are asking? – lulu Apr 29 '21 at 14:43
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    Induction works when $k \in \mathbb{N}$, not for $k \in \mathbb{R}$. – zkutch Apr 29 '21 at 14:44
  • For this particular problem, I'd look at the function $f(x)=\left(\frac 32\right)^x$ and note that $f(0)=1$ but $f'(x)$ is strictly positive. – lulu Apr 29 '21 at 14:47
  • guys, i know how to solve the problem. my doubt is why the limit doesnt work. k and k+1 work well but why dont k and k+h work – Dusty_Wanderer Apr 29 '21 at 15:01

1 Answers1

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The proof technique of induction is usually only used for natural numbers (0 inclusive). So I think you would be better off with a deductive proof rather than doing it by induction.

Check out this post for more about why.

devam_04
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