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Question Number 23928 by Tinkutara last updated on 10/Nov/17

$$\mathrm{Find}{}^n{C}_1-\frac{1}{2}{}^n{C}_2+\frac{1}{3}{}^n{C}_3-....+$$$${(-1)}^{n-1}\frac{1}{n}\cdot {}^n{C}_n.$$

Commented by prakash jain last updated on 10/Nov/17

$$\frac{1-{(1-x)}^n}{x}=\frac{1}{x}\left(\underset{i=1}{\sum ^n}{(-1)}^{i-1}{}^n{C}_i{x}^i\right)$$$$\frac{1-{(1-x)}^n}{x}=\left(\underset{i=1}{\sum ^n}{(-1)}^{i-1}{}^n{C}_i{x}^{i-1}\right)$$$${\int }_0^1\frac{1-{(1-x)}^n}{x}dx={\int }_0^1\left(\underset{i=1}{\sum ^n}{(-1)}^{i-1}{}^n{C}_i{x}^{i-1}\right)dx$$$$\mathrm{RHS}=requiredsum$$$$\mathrm{LHS}=result$$

Commented by prakash jain last updated on 10/Nov/17

$$\mathrm{Yes}.\mathrm{Result}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{integral}$$$$\mathrm{but}\mathrm{it}\mathrm{may}\mathrm{not}\mathrm{be}\mathrm{the}\mathrm{answer}\mathrm{you}$$$$\mathrm{are}\mathrm{looking}\mathrm{for}.$$$$$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{answer}?\mathrm{We}\mathrm{can}\mathrm{use}\mathrm{another}$$$$\mathrm{approach}\mathrm{to}\mathrm{reduce}\mathrm{the}\mathrm{expression}$$$$\mathrm{to}\mathrm{required}\mathrm{form}.$$

Commented by prakash jain last updated on 11/Nov/17

$$\frac{1-{(1-x)}^n}{x}=1+(1-x)+{(1-x)}^2+..+{(1-x)}^{n-1}$$$$\mathrm{RHS}\mathrm{is}\mathrm{GP}witha=1,r=1-x=\frac{1(1-{(1-x)}^n)}{1-(1-x)}$$$${\int }_0^1\frac{1-{(1-x)}^n}{x}dx$$$$={[x-\frac{{(1-x)}^2}{2}-\frac{{(1-x)}^3}{3}-...-\frac{{(1-x)}^n}{n}]}_0^1$$$$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

Commented by Tinkutara last updated on 11/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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