prove-by-mathematical-induction-that-for-all-positive-integers-n-r-1-n-r-r-1-n-3-n-1-n-2-

Question Number 69762 by Rio Michael last updated on 27/Sep/19

prove by mathematical induction, that for all positive integers n, Σ_(r=1) ^n r(r + 1) = (n/3)(n + 1)( n + 2)

$${prove}\:{by}\:{mathematical}\:{induction},\:{that}\:{for}\:{all}\:{positive}\:{integers}\:{n}, \\ $$$$\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}\left({r}\:+\:\mathrm{1}\right)\:=\:\frac{{n}}{\mathrm{3}}\left({n}\:+\:\mathrm{1}\right)\left(\:{n}\:+\:\mathrm{2}\right) \\ $$

Commented by mathmax by abdo last updated on 27/Sep/19

1×2+2×3+3×4+....+n(n+1)=((n(n+1)(n+2))/3) (★) n=1 we get 1×2=((1×2×3)/3) the relation is true let suppose (★) true we have 1×2+2×3+....+(n+1)(n+2)=1×2 +2×3+...+n(n+1) +(n+1)(n+2) =((n(n+1)(n+2))/3) +(n+1)(n+2) =(n+1)(n+2)((n/3)+1) =(((n+1)(n+2)(n+3))/3) so the relstion is true at term(n+1).

$$\mathrm{1}×\mathrm{2}+\mathrm{2}×\mathrm{3}+\mathrm{3}×\mathrm{4}+….+{n}\left({n}+\mathrm{1}\right)=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}\:\:\:\left(\bigstar\right) \\ $$$${n}=\mathrm{1}\:\:\:\:{we}\:{get}\:\mathrm{1}×\mathrm{2}=\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}}{\mathrm{3}}\:\:{the}\:{relation}\:{is}\:{true} \\ $$$${let}\:{suppose}\:\left(\bigstar\right)\:{true}\:\:{we}\:{have} \\ $$$$\mathrm{1}×\mathrm{2}+\mathrm{2}×\mathrm{3}+….+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)=\mathrm{1}×\mathrm{2}\:+\mathrm{2}×\mathrm{3}+…+{n}\left({n}+\mathrm{1}\right) \\ $$$$+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}}\:+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left(\frac{{n}}{\mathrm{3}}+\mathrm{1}\right)\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}{\mathrm{3}} \\ $$$${so}\:{the}\:{relstion}\:{is}\:{true}\:{at}\:{term}\left({n}+\mathrm{1}\right). \\ $$

Answered by $@ty@m123 last updated on 27/Sep/19

P(k+1): Σ_(r=1) ^(k+1) r(r+1)=(k/3)(k+ 1)( k + 2)+( k + 1)( k + 2) =( k + 1)( k + 2){(k/3)+1} =( k + 1)( k + 2)(((k+3)/3)) =((k+1)/3)( k + 2)( k + 3) ⇒P(k+1) is true.

$${P}\left({k}+\mathrm{1}\right): \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}{r}\left({r}+\mathrm{1}\right)=\frac{{k}}{\mathrm{3}}\left({k}+\:\mathrm{1}\right)\left(\:{k}\:+\:\mathrm{2}\right)+\left(\:{k}\:+\:\mathrm{1}\right)\left(\:{k}\:+\:\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\:{k}\:+\:\mathrm{1}\right)\left(\:{k}\:+\:\mathrm{2}\right)\left\{\frac{{k}}{\mathrm{3}}+\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\:{k}\:+\:\mathrm{1}\right)\left(\:{k}\:+\:\mathrm{2}\right)\left(\frac{{k}+\mathrm{3}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{k}+\mathrm{1}}{\mathrm{3}}\left(\:{k}\:+\:\mathrm{2}\right)\left(\:{k}\:+\:\mathrm{3}\right) \\ $$$$\Rightarrow{P}\left({k}+\mathrm{1}\right)\:{is}\:{true}. \\ $$

Commented by Rio Michael last updated on 27/Sep/19