Mathematical Induction Questions Summation and Inequality

Question

For all three questions use the principle of mathematical induction to show that

for all integers $n≥3$

$(1-\frac23)(1-\frac23)(1-\frac23)+...+(1-\frac2n)= \frac2{n(n+1)}$

$2*1!+5*2!+10*3!+...+(n^2+1)n! = n(n+1)!$
For all integers $n≥2$

$12^n>7^n+5^n$

I am able to get through until the prove true for n+1 step. For all questions, I am having difficulty with the proofs. For 1, I am stuck on expansion of the fractions and it does not get the right answer. In 2,I do not understand where to go after I have shown what to get for prove true for n+1. I am completely stuck on question 3. — Wilson Huynh, Nov 29 '15 at 10:59
I think there is a mistake in the expression of 1. $(1-\frac23)(1-\frac23)(1-\frac23)\times...\times(1-\frac2n)= \frac2{n(n+1)}$, not +...+ — Claudeh5, Nov 29 '15 at 11:12
I wouldn't know if there is a mistake in the expression but I had a feeling the question was like that. I am unsure of where to go next. — Wilson Huynh, Nov 29 '15 at 11:17

score 1 · Accepted Answer · answered Nov 29 '15 at 11:24

I can't comment, so I post my answers here.

I think the question should be $(1-\frac23)(1-\frac24)(1-\frac25)\cdots(1-\frac2n)= \frac2{(n-1)n}$

For example, when $n=5$, $(1-\frac23)(1-\frac24)(1-\frac25)=\frac13\frac24\frac35=\frac2{4\cdot5}$

I think you can do it without induction.

Assume $(1-\frac23)(1-\frac24)(1-\frac25)\cdots(1-\frac2k)= \frac2{(k-1)k}$. $(1-\frac23)(1-\frac24)(1-\frac25)\cdots(1-\frac2{k+1})=\frac2{(k-1)k}(1-\frac2{k+1})=\frac2{(k-1)k}\frac{k-1}{k+1}=\frac2{k(k+1)}$

2.$2*1!+5*2!+10*3!+...+(k^2+1)k! = k(k+1)!$ \begin{split} &\quad2*1!+5*2!+10*3!+...+(k^2+1)k!+((k+1)^2+1)(k+1)!\\ &= k(k+1)!+((k+1)^2+1)(k+1)!\\ &= (k+1)!(k^2+3k+2)\\ &= (k+2)!(k+1) \end{split}

Obvious when n = 2. Assume $12^k>7^k+5^k$. As Element118 says, $7^{k+1}+5^{k+1}<(7+5)(7^k+5^k)=12(7^k+5^k)<12\times12^k=12^{k+1}$.

Mathematical Induction Questions Summation and Inequality

1 Answers1