# Omega

less info
reputation
1516
bio website location age 19 member for 3 years, 2 months seen Mar 8 at 0:40 profile views 170

# 339 Actions

 Mar8 awarded Popular Question Mar4 awarded Notable Question Jan28 awarded Popular Question Jan17 awarded Famous Question Jan13 awarded Yearling Dec5 comment Advice for proving with induction scenarios with multiple chances for using the hypothesis. Thanks. Yeah, I know I asked for this scenario - this was out of curiosity. Dec5 comment Advice for proving with induction scenarios with multiple chances for using the hypothesis. Hm, I'm not very familiar with that, but it sounds interesting - could you elaborate a bit more? Dec5 comment Advice for proving with induction scenarios with multiple chances for using the hypothesis. If you were in my situation, looking at these four options, you'd also pick one by one? Well, I guess that's a relief, kinda lol. Dec5 comment Advice for proving with induction scenarios with multiple chances for using the hypothesis. @HagenvonEitzen: Yes, to choose which of the four terms to replace as my next step. Dec5 asked Advice for proving with induction scenarios with multiple chances for using the hypothesis. Dec5 accepted Finding the explicit formula for the succession $x_0=2, x_{n+1} = 5x_n$ and proving it with induction Dec5 asked Finding the explicit formula for the succession $x_0=2, x_{n+1} = 5x_n$ and proving it with induction Dec5 accepted Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$. Dec5 comment Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$. @user112167: It actually came from an induction exercise. I added it there now. Dec5 revised Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$. added 346 characters in body Dec5 asked Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$. Dec4 comment Proving by induction inequalities that lack the variable on the right side. @SammyBlack: Someone told me that there is no problem with starting from $\frac{1}{n+1}$, because it just means that the succession begins from $\frac{1}{n+1}$ and ends on $\frac{1}{(n+1)+1}$. Is that right? Dec3 accepted Proving by induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \ge 1$ Dec3 asked Proving by induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \ge 1$ Dec3 comment Proving by induction inequalities that lack the variable on the right side. @PraphullaKoushik: Well, it's an induction exercise, so I should use it. I don't recognise that term for inequality (or perhaps I do, but in Spanish) - I will take a look at it, thanks.