Menu Close

Question-100223

Tinku Tara 0 Comments
Pin




Question Number 100223 by DGmichael last updated on 25/Jun/20
Answered by maths mind last updated on 25/Jun/20
A=Σ_(k=0) ^n (((−1)^k C_n ^k )/(k+1)) let f(x)=(1−x)^n ⇒f(x)=Σ_(k=0) ^n C_n ^k (−1)^k x^k ∫_0 ^1 (1−x)^n dx=∫_0 ^1 Σ_(k=0) ^n C_n ^k (−1)^k x^k dx =Σ_(k=0) ^n C_n ^k (−1)^k ∫_0 ^1 x^k dx=ΣC_n ^k (−1)^k .(1/(k+1)) =Σ_(k=0) ^n (((−1)^k C_n ^k )/(k+1))=∫_0 ^1 (1−x)^n dx=(1/(1+n))
$${A}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} }{{k}+\mathrm{1}} \\ $$$${let}\:{f}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{{n}} \Rightarrow{f}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}} {dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} {dx}=\Sigma{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{{k}} }{{k}+\mathrm{1}}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} {dx}=\frac{\mathrm{1}}{\mathrm{1}+{n}} \\ $$
Commented by Ar Brandon last updated on 25/Jun/20
wow amazing ! you quickly recognized the integral.��
Commented by Ar Brandon last updated on 25/Jun/20
Excuse me Sir, Is there any theory for this, or any particular topic that deals with these kind of sums ?
Commented by maths mind last updated on 25/Jun/20
the main idea is newtoon identitie (x+y)^n =Σ_(k=0) ^n C_n ^k x^k y^(n−k) then use it integral or differebtial ∂_x (x+y)^n =∂_x Σ_(k=0) ^n C_n ^k x^k y^(n−k) ⇔n(x+y)^(n−1) =Σ_(k≥1) C_n ^k kx^(k−1) y^(n−k) or (((x+y)^(n+1) )/(n+1))=ΣC_n ^k ((x^(k+1) y^(n−k) )/(k+1)) exempl we want Σ_(k=0) ^n (C_n ^k /((k+1)(k+2)(k+3)))... (1+x)^n =Σ_(k=0) ^n C_n ^k x^k ⇒∫_0 ^t (1+x)^n dx=Σ∫_0 ^t C_n ^k x^k dx ⇒(((1+t)^n −1)/(n+1))=Σ_(k=0) ^n ((C_n ^k t^(k+1) )/(k+1)) ∫_0 ^x (((1+t)^n −1)/((n+1)))dt=Σ_(k=0) ^n C_n ^k ∫_0 ^x (t^(k+1) /((k+1)))dt=ΣC_n ^k (x^(k+2) /((k+1)(k+2))) (((1+x)^(n+2) )/((n+1)(n+2)))−(x/(n+1))=ΣC_n ^k (x^(k+2) /((k+1)(k+2))) ⇒∫_0 ^y (((1+x)^(n+2) )/((n+1)(n+2)))−(x/(n+1))=ΣC_n ^k ∫_0 ^y (x^(k+2) /((k+1)(k+2))) ⇔(((1+y)^(n+3) )/((n+1)(n+2)(n+3)))−(1/((n+1)(n+2)))−(y^2 /(2(n+1)))=Σ_(k=0) ^n C_n ^k (y^(k+3) /((k+2)(k+1)(k+3))) y=1⇔(2^(n+3) /((n+1)(n+2)(n+3)))−(1/((n+1)(n+2)))−(1/(2(n+1)))=Σ_(k=0) ^n C_n ^k .(1/((k+1)(k+2)(k+3))) we can Show Σ_(k=0) ^n (C_n ^k /(Π_(b=1) ^r (k+b)))=(2^(r+n) /((n+1).....(n+r)))−Σ_(j=1) ^(r−1) (1/((r−j)!Π_(k=1) ^j (n+j)))
$${the}\:{main}\:{idea} \\ $$$${is}\:{newtoon}\:{identitie} \\ $$$$\left({x}+{y}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{k}} {y}^{{n}−{k}} \\ $$$${then}\:{use}\:{it}\:{integral}\:{or}\:{differebtial} \\ $$$$\partial_{{x}} \left({x}+{y}\right)^{{n}} =\partial_{{x}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{k}} {y}^{{n}−{k}} \Leftrightarrow{n}\left({x}+{y}\right)^{{n}−\mathrm{1}} =\underset{{k}\geqslant\mathrm{1}} {\sum}{C}_{{n}} ^{{k}} {kx}^{{k}−\mathrm{1}} {y}^{{n}−{k}} \\ $$$${or}\:\frac{\left({x}+{y}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\Sigma{C}_{{n}} ^{{k}} \frac{{x}^{{k}+\mathrm{1}} {y}^{{n}−{k}} }{{k}+\mathrm{1}} \\ $$$${exempl}\: \\ $$$${we}\:{want} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{n}} ^{{k}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}… \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{{k}} \Rightarrow\int_{\mathrm{0}} ^{{t}} \left(\mathrm{1}+{x}\right)^{{n}} {dx}=\Sigma\int_{\mathrm{0}} ^{{t}} {C}_{{n}} ^{{k}} {x}^{{k}} {dx} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}+{t}\right)^{{n}} −\mathrm{1}}{{n}+\mathrm{1}}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{n}} ^{{k}} {t}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \frac{\left(\mathrm{1}+{t}\right)^{{n}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)}{dt}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \int_{\mathrm{0}} ^{{x}} \frac{{t}^{{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right)}{dt}=\Sigma{C}_{{n}} ^{{k}} \frac{{x}^{{k}+\mathrm{2}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\frac{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{{x}}{{n}+\mathrm{1}}=\Sigma{C}_{{n}} ^{{k}} \frac{{x}^{{k}+\mathrm{2}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{y}} \frac{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}−\frac{{x}}{{n}+\mathrm{1}}=\Sigma{C}_{{n}} ^{{k}} \int_{\mathrm{0}} ^{{y}} \frac{{x}^{{k}+\mathrm{2}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)} \\ $$$$\Leftrightarrow\frac{\left(\mathrm{1}+{y}\right)^{{n}+\mathrm{3}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}\left({n}+\mathrm{1}\right)}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \frac{{y}^{{k}+\mathrm{3}} }{\left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right)\left({k}+\mathrm{3}\right)} \\ $$$${y}=\mathrm{1}\Leftrightarrow\frac{\mathrm{2}^{{n}+\mathrm{3}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} .\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)} \\ $$$${we}\:{can}\:{Show} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{C}_{{n}} ^{{k}} }{\underset{{b}=\mathrm{1}} {\overset{{r}} {\prod}}\left({k}+{b}\right)}=\frac{\mathrm{2}^{{r}+{n}} }{\left({n}+\mathrm{1}\right)…..\left({n}+{r}\right)}−\underset{{j}=\mathrm{1}} {\overset{{r}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left({r}−{j}\right)!\underset{{k}=\mathrm{1}} {\overset{{j}} {\prod}}\left({n}+{j}\right)} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 26/Jun/20
Thanks a lot for your time Sir��
Commented by DGmichael last updated on 26/Jun/20
�� thanks dear sir!

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *