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Question Number 177931 by Spillover last updated on 11/Oct/22

Prove by the principle of   induction that  1.4.7+2.5.8+3.6.9+...n(n+3)(n+6)  =(n/4)(n+1)(n+6)(n+7)

$$\mathrm{Prove}\:\mathrm{by}\:\mathrm{the}\:\mathrm{principle}\:\mathrm{of}\: \\ $$$$\mathrm{induction}\:\mathrm{that} \\ $$$$\mathrm{1}.\mathrm{4}.\mathrm{7}+\mathrm{2}.\mathrm{5}.\mathrm{8}+\mathrm{3}.\mathrm{6}.\mathrm{9}+...\mathrm{n}\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{6}\right) \\ $$$$=\frac{\mathrm{n}}{\mathrm{4}}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{6}\right)\left(\mathrm{n}+\mathrm{7}\right) \\ $$

Answered by Ar Brandon last updated on 11/Oct/22

Test for k=1 , k=2,   assume P_k  is true for n and deduce that it′s true for n+1  P_n : Σ_(k=1) ^n k(k+3)(k+6)=(n/4)(n+1)(n+6)(n+7)  P_(n+1) : Σ_(k=1) ^(n+1) k(k+3)(k+6)=P_n +(n+1)^(th)  term             =(n/4)(n+1)(n+6)(n+7)+(n+1)(n+4)(n+7)             =(n+1)(n+7)[(n/4)(n+6)+n+4]             =(((n+1)(n+7))/4)(n^2 +10n+16)              =(((n+1)(n+7))/4)(n+2)(n+8)              =(((n+1))/4)(n+2)(n+7)(n+8)  ... Conclusion...

$$\mathrm{Test}\:\mathrm{for}\:{k}=\mathrm{1}\:,\:{k}=\mathrm{2}, \\ $$$$\:\mathrm{assume}\:{P}_{{k}} \:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{n}\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{that}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}\:\mathrm{for}\:{n}+\mathrm{1} \\ $$$${P}_{{n}} :\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}+\mathrm{3}\right)\left({k}+\mathrm{6}\right)=\frac{{n}}{\mathrm{4}}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{6}\right)\left({n}+\mathrm{7}\right) \\ $$$${P}_{{n}+\mathrm{1}} :\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}{k}\left({k}+\mathrm{3}\right)\left({k}+\mathrm{6}\right)={P}_{{n}} +\left({n}+\mathrm{1}\right)^{\mathrm{th}} \:\mathrm{term} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{n}}{\mathrm{4}}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{6}\right)\left({n}+\mathrm{7}\right)+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{4}\right)\left({n}+\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{7}\right)\left[\frac{{n}}{\mathrm{4}}\left({n}+\mathrm{6}\right)+{n}+\mathrm{4}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{7}\right)}{\mathrm{4}}\left({n}^{\mathrm{2}} +\mathrm{10}{n}+\mathrm{16}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{7}\right)}{\mathrm{4}}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}+\mathrm{1}\right)}{\mathrm{4}}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{7}\right)\left({n}+\mathrm{8}\right) \\ $$$$...\:\mathrm{Conclusion}... \\ $$

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