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Question Number 111132 by bemath last updated on 02/Sep/20

$$\mathrm{prove}\mathrm{by}\mathrm{mathematical}\mathrm{induction}$$$$\Rightarrow 7^{\mathrm{n}}-(3\mathrm{n}+4)\times 4^{\mathrm{n}-1}\mathrm{divided}\mathrm{by}9$$

Answered by john santu last updated on 02/Sep/20

$$letp\left(n\right)=7^n-(3n+4).4^{n-1}$$$$\left(1\right)p\left(1\right)=7-(3.1+4).4^0=7-7=0\left(true\right)$$$$\left(2\right)assumeforn=k\to p\left(k\right)divisibleby9$$$$wehavep\left(k\right)=7^k-(3k+4).4^{k-1}\equiv 9m,m\in \mathbb{Z}$$$$=7^k-3k.4^{k-1}-4^k\equiv 9m$$$$=7^k-4^k-3k.4^{k-1}\equiv 9m$$$$\left(3\right)wewanttoproveforn=k+1$$$$p(k+1)isdivisibleby9$$$$p(k+1)=7^{k+1}-(3(k+1)+4).4^k$$$$\iff 7.7^k-(3k+4+3).4^k$$$$\iff 7.7^k-(3k+7).4^k$$$$\Rightarrow 7.7^k-3k.4^k-7.4^k$$$$\Rightarrow 7(7^k-4^k)-3k.4^k$$$$\Rightarrow 7(7^k-4^k-3k.4^{k-1})+21k.4^{k-1}-3k.4^k$$$$\Rightarrow 7\left(9m\right)+3k.4^{k-1}(7-4)$$$$\Rightarrow 63m+9k.4^{k-1}$$$$\Rightarrow 9(7m+k.4^{k-1})isdivisibleby9.$$$$q.e.d$$

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