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

$$\mathrm{Prove}\mathrm{by}\mathrm{the}\mathrm{principle}\mathrm{of}$$$$\mathrm{induction}\mathrm{that}$$$$1.4.7+2.5.8+3.6.9+...\mathrm{n}(\mathrm{n}+3)(\mathrm{n}+6)$$$$=\frac{\mathrm{n}}{4}(\mathrm{n}+1)(\mathrm{n}+6)(\mathrm{n}+7)$$

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

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

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