I was recently taught strong induction in class and overall I'm still learning the topic. I am currently stuck on a problem and I don't even know where to begin with it. A starting place or hint would be great.
Proving a summation result using strong induction
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$\begingroup$
summation
induction
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0"Hint : use a technique of exercise 3" ... So we need to have a look at exercise 3 ... This apart, it looks like a "telescopic summation" : try to factorize $j(j+1)(j+2)-(j+1)(j+2)(j+3)$ as a starting point – 2017-02-15
1 Answers
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Fix $k$ and apply induction on $n$. So, let us assume that $$\sum_{j=1}^\color{blue}nj(j+1)(j+2)\dots(j+k-1)=\frac{\color{blue}n(\color{blue}n+1)(\color{blue}n+2)\dots(\color{blue}n+k)}{k+1}.$$ Then $$\begin{align} &\sum_{j=1}^{\color{red}{n+1}}j(j+1)(j+2)\dots(j+k-1)\\ &=[(n+1)(n+2)(n+3)\dots(n+1+k-1)]+\sum_{j=1}^nj(j+1)(j+2)\dots(j+k-1)\\ &=[(n+1)(n+2)(n+3)\dots(n+k)]+\sum_{j=1}^nj(j+1)(j+2)\dots(j+k-1)\\ &=\overbrace{[(n+1)(n+2)(n+3)\dots(n+k)]}+\frac{n\overbrace{(n+1)(n+2)(n+3)\dots(n+k)}}{k+1}\\ &=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[1+\frac{n}{k+1}\right]\\ &=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[\frac{k+1+n}{k+1}\right]\\ &=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[\frac{n+k+1}{k+1}\right]\\ &=\frac{(n+1)(n+2)(n+3)\dots(n+k)(n+k+1)}{k+1}\\ &=\frac{\color{red}{(n+1)}\cdot[\color{red}{(n+1)}+1)]\cdot[\color{red}{(n+1)}+2]\dots[\color{red}{(n+1)}+k]}{k+1} \end{align}$$ Hope this help.
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0Nicely avoids the usual magic telescoping sum. – 2017-02-15
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0@marty cohen I did it correctly? – 2017-02-15
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0Looks OK to me. – 2017-02-15
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0Where did the [1+(n)/(k+1)] come from in the 5th line? – 2017-02-15
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0@Anthony L The common factor is the one that I marked as braces. Factor out it to obtain line 5. – 2017-02-15