Menu Close

Let-1-x-n-r-0-n-C-r-x-r-and-r-0-n-C-r-r-1-k-then-the-value-of-k-is-

Tinku Tara 0 Comments
Pin




Question Number 63942 by gunawan last updated on 11/Jul/19
Let (1+x)^n = Σ_(r=0) ^n C_r x^r and Σ_(r=0) ^n (C_r /(r+1)) = k, then the value of k is
$$\mathrm{Let}\:\left(\mathrm{1}+{x}\right)^{{n}} =\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:{C}_{{r}} \:{x}^{{r}} \:\mathrm{and}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{{C}_{{r}} }{{r}+\mathrm{1}}\:=\:{k}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is} \\ $$
Commented by mathmax by abdo last updated on 11/Jul/19
(1+x)^n =Σ_(r=0) ^n C_n ^r x^r ⇒C_r =C_n ^r ⇒ Σ_(r=0) ^n (C_r /(r+1)) =Σ_(r=0) ^n (C_n ^r /(r+1)) let p(x) =Σ_(r=0) ^n (C_n ^r /(r+1)) x^(r+1) ⇒p^′ (x) =Σ_(r=0) ^n C_n ^r x^r =(x+1)^n ⇒ p(x)=∫ (x+1)^n dx +c =(1/(n+1))(x+1)^(n+1) +c p(0)=0=(1/(n+1)) +c ⇒c=−(1/(n+1)) ⇒p(x) =(((x+1)^(n+1) −1)/(n+1)) ⇒ Σ_(r=0) ^∞ (C_r /(r+1)) =p(1) =((2^(n+1) −1)/(n+1)) ⇒k =((2^(n+1) −1)/(n+1)) .
$$\left(\mathrm{1}+{x}\right)^{{n}} \:=\sum_{{r}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{r}} \:{x}^{{r}} \:\Rightarrow{C}_{{r}} ={C}_{{n}} ^{{r}} \:\Rightarrow \\ $$$$\sum_{{r}=\mathrm{0}} ^{{n}} \:\frac{{C}_{{r}} }{{r}+\mathrm{1}}\:=\sum_{{r}=\mathrm{0}} ^{{n}} \:\:\frac{{C}_{{n}} ^{{r}} }{{r}+\mathrm{1}} \\ $$$${let}\:{p}\left({x}\right)\:=\sum_{{r}=\mathrm{0}} ^{{n}} \:\frac{{C}_{{n}} ^{{r}} }{{r}+\mathrm{1}}\:{x}^{{r}+\mathrm{1}} \:\Rightarrow{p}^{'} \left({x}\right)\:=\sum_{{r}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{r}} \:{x}^{{r}} \:\:=\left({x}+\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${p}\left({x}\right)=\int\:\left({x}+\mathrm{1}\right)^{{n}} {dx}\:+{c}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:+{c} \\ $$$${p}\left(\mathrm{0}\right)=\mathrm{0}=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+{c}\:\Rightarrow{c}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow{p}\left({x}\right)\:=\frac{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{r}=\mathrm{0}} ^{\infty} \:\frac{{C}_{{r}} }{{r}+\mathrm{1}}\:={p}\left(\mathrm{1}\right)\:=\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow{k}\:=\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:. \\ $$$$ \\ $$
Answered by mr W last updated on 11/Jul/19
(1+x)^n = Σ_(r=0) ^n C_r x^r ∫_0 ^x (1+x)^n dx= Σ_(r=0) ^n ∫_0 ^x C_r x^r dx (1/(n+1))(1+x)^(n+1) −(1/(n+1))= Σ_(r=0) ^n (1/(r+1)) C_r x^(r+1) let x=1: ((2^(n+1) −1)/(n+1))= Σ_(r=0) ^n (1/(r+1)) C_r =k ⇒k=((2^(n+1) −1)/(n+1))
$$\left(\mathrm{1}+{x}\right)^{{n}} =\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:{C}_{{r}} \:{x}^{{r}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}+{x}\right)^{{n}} {dx}=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\int_{\mathrm{0}} ^{{x}} \:{C}_{{r}} \:{x}^{{r}} {dx} \\ $$$$\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}+\mathrm{1}}\:{C}_{{r}} \:{x}^{{r}+\mathrm{1}} \\ $$$${let}\:{x}=\mathrm{1}: \\ $$$$\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}=\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{r}+\mathrm{1}}\:{C}_{{r}} ={k}\: \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}} \\ $$

Pin

Leave a Reply

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