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Question Number 23928 by Tinkutara last updated on 10/Nov/17
Find^n C_1 − (1/2)^n C_2 + (1/3)^n C_3 − .... + (−1)^(n−1) (1/n) ∙^n C_n .
$$\mathrm{Find}\:^{{n}} {C}_{\mathrm{1}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}\:^{{n}} {C}_{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{3}}\:^{{n}} {C}_{\mathrm{3}} \:−\:….\:+ \\ $$$$\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{n}}\:\centerdot\:^{{n}} {C}_{{n}} . \\ $$
Commented by prakash jain last updated on 10/Nov/17
((1−(1−x)^n )/x)=(1/x)(Σ_(i=1) ^n (−1)^(i−1) ^n C_i x^i ) ((1−(1−x)^n )/x)=(Σ_(i=1) ^n (−1)^(i−1) ^n C_i x^(i−1) ) ∫_0 ^1 ((1−(1−x)^n )/x)dx=∫_0 ^1 (Σ_(i=1) ^n (−1)^(i−1) ^n C_i x^(i−1) )dx RHS = required sum LHS=result
$$\frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}=\frac{\mathrm{1}}{{x}}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(−\mathrm{1}\right)^{{i}−\mathrm{1}} \:^{{n}} {C}_{{i}} {x}^{{i}} \right) \\ $$$$\frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}=\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(−\mathrm{1}\right)^{{i}−\mathrm{1}} \:^{{n}} {C}_{{i}} {x}^{{i}−\mathrm{1}} \right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(−\mathrm{1}\right)^{{i}−\mathrm{1}} \:^{{n}} {C}_{{i}} {x}^{{i}−\mathrm{1}} \right){dx} \\ $$$$\mathrm{RHS}\:=\:{required}\:{sum} \\ $$$$\mathrm{LHS}={result} \\ $$
Commented by prakash jain last updated on 10/Nov/17
Yes. Result is the value of integral but it may not be the answer you are looking for. What is the answer? We can use another approach to reduce the expression to required form.
$$\mathrm{Yes}.\:\mathrm{Result}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{integral} \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{may}\:\mathrm{not}\:\mathrm{be}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{you} \\ $$$$\mathrm{are}\:\mathrm{looking}\:\mathrm{for}. \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}?\:\mathrm{We}\:\mathrm{can}\:\mathrm{use}\:\mathrm{another} \\ $$$$\mathrm{approach}\:\mathrm{to}\:\mathrm{reduce}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{to}\:\mathrm{required}\:\mathrm{form}. \\ $$
Commented by prakash jain last updated on 11/Nov/17
((1−(1−x)^n )/x)=1+(1−x)+(1−x)^2 +..+(1−x)^(n−1) RHS is GP with a=1,r=1−x=((1(1−(1−x)^n ))/(1−(1−x))) ∫_0 ^1 ((1−(1−x)^n )/x)dx =[x−(((1−x)^2 )/2) − (((1−x)^3 )/3) −...−(((1−x)^n )/n)]_0 ^1 =1+(1/2)+(1/3)+...+(1/n)
$$\frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}=\mathrm{1}+\left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +..+\left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}} \\ $$$$\mathrm{RHS}\:\mathrm{is}\:\mathrm{GP}\:{with}\:{a}=\mathrm{1},{r}=\mathrm{1}−{x}=\frac{\mathrm{1}\left(\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} \right)}{\mathrm{1}−\left(\mathrm{1}−{x}\right)} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{{n}} }{{x}}{dx} \\ $$$$=\left[{x}−\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:−\:\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }{\mathrm{3}}\:−…−\frac{\left(\mathrm{1}−{x}\right)^{{n}} }{{n}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}} \\ $$
Commented by Tinkutara last updated on 11/Nov/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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