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

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

$${(1+x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{x}^k$$$${(1+x)}^n{x}^2=\underset{k=0}{\sum ^n}{C}_k^n{x}^{k+2}$$$${\int }_0^t{(1+x)}^n{x}^{2}dx=\underset{k=0}{\sum ^n}{C}_k^n{\int }_0^t{x}^{k+2}dx$$$${\int }_0^t{(1+x)}^n{x}^{2}dx=\underset{k=0}{\sum ^n}\frac{C_k^n{t}^{k+3}}{(k+3)}$$$$\frac{1}{t^3}{\int }_0^t{(1+x)}^n{x}^{2}dx=\underset{k=0}{\sum ^n}\frac{C_k^n{t}^k}{(k+3)}$$$$lett=\frac{1}{n}$$$$n^3{\int }_0^{\frac{1}{n}}{(1+x)}^n{x}^{2}dx=\underset{k=0}{\sum ^n}\frac{C_k^n}{(k+3)n^k}=\mathrm{\Phi }$$$$\mathrm{\Phi }=n^3{\int }_0^{\frac{1}{n}}{(1+x)}^n{x}^{2}dx$$$$\mathrm{\Phi }=\frac{(n+1)(n^2+n+2){(1+\frac{1}{n})}^n-2n^3}{(n+1)(n+2)(n+3)}$$$$\mathrm{\Phi }=\frac{(1+\frac{1}{n})(1+\frac{1}{n}+\frac{2}{n^2}){(1+\frac{1}{n})}^n-2}{(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})}$$$$\underset{n\to \mathrm{\infty }}{\lim }\mathrm{\Phi }=e-2✓$$

Commented by infinityaction last updated on 16/Sep/22

$$thankyousir$$

Commented by Tawa11 last updated on 18/Sep/22

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

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