Question-176199

Question Number 176199 by infinityaction last updated on 14/Sep/22

Answered by mr W last updated on 15/Sep/22

(1+x)^n =Σ_(k=0) ^n C_k ^n x^k (1+x)^n x^2 =Σ_(k=0) ^n C_k ^n x^(k+2) ∫_0 ^t (1+x)^n x^2 dx=Σ_(k=0) ^n C_k ^n ∫_0 ^t x^(k+2) dx ∫_0 ^t (1+x)^n x^2 dx=Σ_(k=0) ^n ((C_k ^n t^(k+3) )/((k+3))) (1/t^3 )∫_0 ^t (1+x)^n x^2 dx=Σ_(k=0) ^n ((C_k ^n t^k )/((k+3))) let t=(1/n) n^3 ∫_0 ^(1/n) (1+x)^n x^2 dx=Σ_(k=0) ^n (C_k ^n /((k+3)n^k ))=Φ Φ=n^3 ∫_0 ^(1/n) (1+x)^n x^2 dx Φ=(((n+1)(n^2 +n+2)(1+(1/n))^n −2n^3 )/((n+1)(n+2)(n+3))) Φ=(((1+(1/n))(1+(1/n)+(2/n^2 ))(1+(1/n))^n −2)/((1+(1/n))(1+(2/n))(1+(3/n)))) lim_(n→∞) Φ=e−2 ✓

$$\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {x}^{{k}} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {x}^{{k}+\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} \int_{\mathrm{0}} ^{{t}} {x}^{{k}+\mathrm{2}} {dx} \\ $$$$\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{k}} ^{{n}} {t}^{{k}+\mathrm{3}} }{\left({k}+\mathrm{3}\right)} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{k}} ^{{n}} {t}^{{k}} }{\left({k}+\mathrm{3}\right)} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{{n}} \\ $$$${n}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} {dx}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{k}} ^{{n}} }{\left({k}+\mathrm{3}\right){n}^{{k}} }=\Phi \\ $$$$\Phi={n}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}+{x}\right)^{{n}} {x}^{\mathrm{2}} {dx} \\ $$$$\Phi=\frac{\left({n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +{n}+\mathrm{2}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} −\mathrm{2}{n}^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$\Phi=\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} −\mathrm{2}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\Phi={e}−\mathrm{2}\:\checkmark \\ $$

Commented by infinityaction last updated on 16/Sep/22

$${thank}\:{you}\:{sir} \\ $$

Commented by Tawa11 last updated on 18/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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