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Question Number 129569 by zakirullah last updated on 16/Jan/21
Qi. using binomial theorem, prove that (3^(2n+2 ) −8n−9) is divisible by 64, where n is positive integer. Qii. By mathematical induction; (cosα+(n−1)β) = cos(α+(((n−1)β)/2))×((sin(nβ/2))/(sin(β/2)))
$$\mathrm{Qi}.\:\mathrm{using}\:\mathrm{binomial}\:\mathrm{theorem},\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{3}^{\mathrm{2n}+\mathrm{2}\:} −\mathrm{8n}−\mathrm{9}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{64},\:\mathrm{where} \\ $$$$\:\mathrm{n}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer}. \\ $$$$\mathrm{Qii}.\:\:\mathrm{By}\:\mathrm{mathematical}\:\mathrm{induction};\: \\ $$$$\:\:\left(\mathrm{cos}\alpha+\left(\mathrm{n}−\mathrm{1}\right)\beta\right)\:=\:\mathrm{cos}\left(\alpha+\frac{\left(\mathrm{n}−\mathrm{1}\right)\beta}{\mathrm{2}}\right)×\frac{\mathrm{sin}\left(\mathrm{n}\beta/\mathrm{2}\right)}{\mathrm{sin}\left(\beta/\mathrm{2}\right)} \\ $$
Answered by Dwaipayan Shikari last updated on 16/Jan/21
cosα+cos(α+β)+cos(α+2β)+...+cos(α+(n−1)β) =(1/(2sin(β/2)))(sin(α+(β/2))−sin(α−(β/2))+sin(α+((3β)/2))−sin(α+(β/2))+...sin(α+nβ−(β/2))−sin(α+nβ−((3β)/2))) =(1/(2sin(β/2)))(sin(α+nβ−(β/2))−sin(α−(β/2))) =(1/(sin(β/2)))(cos(α+(((n−1)β)/2))sin(((nβ)/2)))
$${cos}\alpha+{cos}\left(\alpha+\beta\right)+{cos}\left(\alpha+\mathrm{2}\beta\right)+…+{cos}\left(\alpha+\left({n}−\mathrm{1}\right)\beta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{\beta}{\mathrm{2}}}\left({sin}\left(\alpha+\frac{\beta}{\mathrm{2}}\right)−{sin}\left(\alpha−\frac{\beta}{\mathrm{2}}\right)+{sin}\left(\alpha+\frac{\mathrm{3}\beta}{\mathrm{2}}\right)−{sin}\left(\alpha+\frac{\beta}{\mathrm{2}}\right)+…{sin}\left(\alpha+{n}\beta−\frac{\beta}{\mathrm{2}}\right)−{sin}\left(\alpha+{n}\beta−\frac{\mathrm{3}\beta}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{\beta}{\mathrm{2}}}\left({sin}\left(\alpha+{n}\beta−\frac{\beta}{\mathrm{2}}\right)−{sin}\left(\alpha−\frac{\beta}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{{sin}\frac{\beta}{\mathrm{2}}}\left({cos}\left(\alpha+\frac{\left({n}−\mathrm{1}\right)\beta}{\mathrm{2}}\right){sin}\left(\frac{{n}\beta}{\mathrm{2}}\right)\right) \\ $$
Commented by zakirullah last updated on 16/Jan/21
thanks sir also find Qi?
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{also}\:\mathrm{find}\:\mathrm{Qi}? \\ $$
Answered by talminator2856791 last updated on 16/Jan/21
3^(2n+2) −8n−9 = 9^(n+1) −8n−9 i. (n+1)^k ≡ 1 (mod n) , k ∈ N ⇒ 9^(n+1) = (8+1)^(n+1) ≡ 1 (mod 8) 9^(n+1) −9 = 9(9^n −1) = 9∙8Σ_(k=0) ^(n−1) 9^k from i., deduce Σ_(k=0) ^(n−1) 9^k ≡ n (mod 8) ⇒ 9∙8Σ_(k=0) ^(n−1) 9^k ≡ 9∙8∙n (mod 8) 9∙8Σ_(k=0) ^(n−1) 9^k ≡ 0 (mod 8) 9∙8Σ_(k=0) ^(n−1) 9^k = 8∙8Σ_(k=0) ^(n−1) 9^k + 8∙Σ_(k=0) ^(n−1) 9^k 8Σ_(k=0) ^(n−1) 9^k ≡ 0 (mod 8) ⇒ 8Σ_(k=0) ^(n−1) 9^k ≡ 8n (mod 64) ⇒ 3^(2n+2) −9 ≡ 8n (mod 64) 3^(2n+2) −8n−9 ≡ 0 (mod 64)
$$\: \\ $$$$\:\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{8}{n}−\mathrm{9}\:=\:\mathrm{9}^{{n}+\mathrm{1}} −\mathrm{8}{n}−\mathrm{9} \\ $$$$\:\mathrm{i}.\:\:\left({n}+\mathrm{1}\right)^{{k}} \:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:{n}\right)\:\:,\:\:\:\:\:{k}\:\in\:\mathbb{N}\: \\ $$$$\:\Rightarrow\:\mathrm{9}^{{n}+\mathrm{1}} \:=\:\left(\mathrm{8}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\: \\ $$$$\:\mathrm{9}^{{n}+\mathrm{1}} −\mathrm{9}\:=\:\mathrm{9}\left(\mathrm{9}^{{n}} −\mathrm{1}\right)\:=\:\mathrm{9}\centerdot\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \: \\ $$$$\:\mathrm{from}\:\mathrm{i}.,\:\mathrm{deduce}\:\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \:\equiv\:{n}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\Rightarrow\:\mathrm{9}\centerdot\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \:\equiv\:\mathrm{9}\centerdot\mathrm{8}\centerdot{n}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{9}\centerdot\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \:\equiv\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{9}\centerdot\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \:=\:\mathrm{8}\centerdot\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \:+\:\mathrm{8}\centerdot\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{9}^{{k}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\mathrm{9}^{{k}} \:\equiv\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{8}\right) \\ $$$$\:\:\Rightarrow\:\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\mathrm{9}^{{k}} \:\equiv\:\mathrm{8}{n}\:\left(\mathrm{mod}\:\mathrm{64}\right) \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\Rightarrow\:\:\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{9}\:\equiv\:\mathrm{8}{n}\:\left(\mathrm{mod}\:\mathrm{64}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{8}{n}−\mathrm{9}\:\equiv\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{64}\right) \\ $$$$\: \\ $$
Commented by zakirullah last updated on 16/Jan/21
you are the great man.
$$\mathrm{you}\:\mathrm{are}\:\mathrm{the}\:\mathrm{great}\:\mathrm{man}. \\ $$

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