Question-202765

Question Number 202765 by Mastermind last updated on 02/Jan/24

Answered by aleks041103 last updated on 02/Jan/24

((x+x^2 +...+x^n −n)/(x−1))= =Σ_(k=1) ^n ((x^k −1)/(x−1))= =Σ_(k=1) ^n Σ_(s=0) ^(k−1) x^s ⇒lim_(x→1) ((x+x^2 +...+x^n −n)/(x−1))=lim_(x→1) Σ_(k=1) ^n Σ_(s=0) ^(k−1) x^s = =Σ_(k=1) ^n Σ_(s=0) ^(k−1) (1)=Σ_(k=1) ^n k=((n(n+1))/2) ⇒((n(n+1))/2)=820 ⇒n^2 +n−1640=0 n_(1,2) =((−1±(√(1+4.1640)))/2)∈N ⇒n=(((√(6561))−1)/2)=40 ⇒n=40

$$\frac{{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} −{n}}{{x}−\mathrm{1}}= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} −\mathrm{1}}{{x}−\mathrm{1}}= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{s}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{x}^{{s}} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{1}} {{lim}}\frac{{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} −{n}}{{x}−\mathrm{1}}=\underset{{x}\rightarrow\mathrm{1}} {{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{s}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}{x}^{{s}} = \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{s}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\left(\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{820} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +{n}−\mathrm{1640}=\mathrm{0} \\ $$$${n}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}.\mathrm{1640}}}{\mathrm{2}}\in\mathbb{N} \\ $$$$\Rightarrow{n}=\frac{\sqrt{\mathrm{6561}}−\mathrm{1}}{\mathrm{2}}=\mathrm{40} \\ $$$$\Rightarrow{n}=\mathrm{40} \\ $$

Answered by ajfour last updated on 02/Jan/24

(1) 40 ((x(x^n −1))/(x−1))=820(x−1)+n let x=1+h(→0) ⇒ x−1=h x^(n+1) −x=820h^2 +nh (1+h)^(n+1) −(1+h)=820h^2 +nh ⇒ {1+(n+1)h+(((n+1)nh^2 )/2)+...} −(1+h)=820h^2 +nh ⇒n(n+1)=2×820 (n+(1/2))^2 =1640+(1/4)=((6561)/4) 2n+1=(√(6561))=81 n=40

$$\left(\mathrm{1}\right)\:\mathrm{40} \\ $$$$\frac{{x}\left({x}^{{n}} −\mathrm{1}\right)}{{x}−\mathrm{1}}=\mathrm{820}\left({x}−\mathrm{1}\right)+{n} \\ $$$${let}\:\:{x}=\mathrm{1}+{h}\left(\rightarrow\mathrm{0}\right)\:\:\:\Rightarrow\:\:{x}−\mathrm{1}={h} \\ $$$${x}^{{n}+\mathrm{1}} −{x}=\mathrm{820}{h}^{\mathrm{2}} +{nh} \\ $$$$\left(\mathrm{1}+{h}\right)^{{n}+\mathrm{1}} −\left(\mathrm{1}+{h}\right)=\mathrm{820}{h}^{\mathrm{2}} +{nh} \\ $$$$\Rightarrow \\ $$$$\left\{\mathrm{1}+\left({n}+\mathrm{1}\right){h}+\frac{\left({n}+\mathrm{1}\right){nh}^{\mathrm{2}} }{\mathrm{2}}+…\right\} \\ $$$$\:\:\:\:−\left(\mathrm{1}+{h}\right)=\mathrm{820}{h}^{\mathrm{2}} +{nh} \\ $$$$\Rightarrow{n}\left({n}+\mathrm{1}\right)=\mathrm{2}×\mathrm{820} \\ $$$$\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1640}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{6561}}{\mathrm{4}} \\ $$$$\mathrm{2}{n}+\mathrm{1}=\sqrt{\mathrm{6561}}=\mathrm{81} \\ $$$${n}=\mathrm{40} \\ $$

Answered by manxsol last updated on 02/Jan/24

(0/0)⇒L′Hospital lim_(x→1) ((1+2x+3x^2 +.....nx^(n−1) )/1) =((n(n+1))/2)=820 n(n+1)=41×20×2=40×41 n=40

$$\frac{\mathrm{0}}{\mathrm{0}}\Rightarrow{L}'{Hospital} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{1}} {{lim}}\frac{\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +…..{nx}^{{n}−\mathrm{1}} }{\mathrm{1}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{820} \\ $$$${n}\left({n}+\mathrm{1}\right)=\mathrm{41}×\mathrm{20}×\mathrm{2}=\mathrm{40}×\mathrm{41} \\ $$$${n}=\mathrm{40} \\ $$

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