Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 143461 by Willson last updated on 14/Jun/21

$$\mathrm{Prove}\mathrm{that}:\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }$$$$\mathrm{a}.\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{C}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}}$$$$\mathrm{b}.\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{C}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}=\frac{4^{\mathrm{n}}}{(2\mathrm{n}+1)C_{2\mathrm{n}}^{\mathrm{n}}}$$

Answered by Dwaipayan Shikari last updated on 14/Jun/21

$$\underset{k=1}{\sum ^n}{C}_n^k{x}^k={(1+x)}^k-1$$$$\Rightarrow \underset{k=1}{\sum ^n}{C}_n^k{x}^{k-1}=\frac{{(1+x)}^n-1}{x}\Rightarrow \underset{k=1}{\sum ^n}{C}_n^k\frac{{(-1)}^k}{k}={\int }_0^{-1}\frac{{(1+x)}^n-1}{x}dx$$$$={\int }_0^1\frac{{(1-u)}^n-1}{u}du=-{\int }_0^1\frac{1-u^n}{1-u}du=-H_n$$

Answered by mathmax by abdo last updated on 14/Jun/21

$$1)\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow {\mathrm{p}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}$$$$=\frac{1}{\mathrm{x}}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\frac{1}{\mathrm{x}}(\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-1)$$$$=\frac{1}{\mathrm{x}}({(1-\mathrm{x})}^{\mathrm{n}}-1)\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{{(1-\mathrm{t})}^{\mathrm{n}}-1}{\mathrm{t}}\mathrm{dt}+\mathrm{K}$$$$\mathrm{K}=\mathrm{p}\left(0\right)=0\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{(1-\mathrm{t}-1)(1+(1-\mathrm{t})+{(1-\mathrm{t})}^2+....+{(1-\mathrm{t})}^{\mathrm{n}-1})}{\mathrm{t}}\mathrm{dt}$$$$=-{\int }_0^{\mathrm{x}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{(1-\mathrm{t})}^{\mathrm{k}}\mathrm{dt}=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\left[\frac{1}{\mathrm{k}+1}{(1-\mathrm{t})}^{\mathrm{k}+1}\right]}_0^{\mathrm{x}}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{1}{\mathrm{k}+1}({(1-\mathrm{x})}^{\mathrm{k}+1}-1)=\mathrm{p}\left(\mathrm{x}\right)\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=\mathrm{p}\left(1\right)=-\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{1}{\mathrm{k}+1}=-\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}=-{\mathrm{H}}_{\mathrm{n}}$$

Answered by mathmax by abdo last updated on 15/Jun/21

$$2)\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}{\mathrm{x}}^{2\mathrm{k}+1}\Rightarrow$$$${\mathrm{p}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\left({\mathrm{x}}^2\right)}^{\mathrm{k}}={(1-{\mathrm{x}}^2)}^{\mathrm{n}}-1\Rightarrow$$$$\mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}({(1-{\mathrm{t}}^2)}^{\mathrm{n}}-1)\mathrm{dt}+\mathrm{K}$$$$\mathrm{k}=\mathrm{p}\left(0\right)=0\Rightarrow \mathrm{p}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}{(1-{\mathrm{t}}^2)}^{\mathrm{n}}\mathrm{dt}-\mathrm{x}$$$$\Rightarrow \mathrm{p}\left(1\right)={\int }_0^1{(1-{\mathrm{t}}^2)}^{\mathrm{n}}\mathrm{dt}-1$$$${\int }_0^1{(1-{\mathrm{t}}^2)}^{\mathrm{n}}\mathrm{dt}{=}_{\mathrm{t}=\mathrm{sinx}}{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}}\mathrm{x}\mathrm{cosx}\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\mathrm{xdx}={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\mathrm{x}{\sin }^{\mathrm{o}}\mathrm{x}\mathrm{dx}$$$$2\mathrm{p}-1=2\mathrm{n}+1\Rightarrow 2\mathrm{p}=2\mathrm{n}+2\Rightarrow \mathrm{p}=\mathrm{n}+1$$$$2\mathrm{q}-1=0\Rightarrow \mathrm{q}=\frac{1}{2}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\mathrm{xdx}={\int }_0^{\frac{\pi }{2}}{\cos }^{2(\mathrm{n}+1)-1}{\sin }^{2\frac{1}{2}-1}\mathrm{xdx}$$$$=\frac{1}{2}\mathrm{B}(\mathrm{n}+1,\frac{1}{2})=\frac{1}{2}\frac{\mathrm{\Gamma }(\mathrm{n}+1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}=\frac{\mathrm{n}!\sqrt{\pi }}{2\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}$$$$..\mathrm{be}\mathrm{continued}....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com