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1-Find-the-sum-s-n-1-2x-3x-2-4x-3-nx-n-1-Hence-or-otherwise-find-the-sum-k-1-n-k-2-k-2-Simplify-the-following-i-r-0-n-2r-1-2n-ii-r-0-n-1-r-r-r-n-i

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Question Number 37991 by Fawomath last updated on 20/Jun/18
1. Find the sum s_n =1+2x+3x^2 +4x^3 +...+nx^(n−1) Hence,or otherwise, find the sum Σ_(k=1) ^n k.2^k 2. Simplify the following i. Σ_(r=0) ^n (_(2r−1) ^(2n) ) ii.Σ_(r=0) ^n (−1)^r r(_r ^n ) iii.Σ_(r=0) ^n (−1)^r (1/(r+1))(_r ^n ) iv.Σ_(r=0) ^n (_(2r) ^(2n) ) v.Σ_(r=0) ^n (−1)^r (_(n−r) ^(n+1) ) 3.Find the sum Σ_(r=0) ^(n−k) (_k ^(n−r) ), where k=0,1,2,3,...,n
$$\mathrm{1}.\:{Find}\:{the}\:{sum} \\ $$$$\:\:\:\:{s}_{{n}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +…+{nx}^{{n}−\mathrm{1}} \\ $$$${Hence},{or}\:{otherwise},\:{find}\:{the}\:{sum} \\ $$$$\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}.\mathrm{2}^{{k}} \\ $$$$\mathrm{2}.\:{Simplify}\:{the}\:{following} \\ $$$${i}.\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{2}{r}−\mathrm{1}} ^{\mathrm{2}{n}} \right) \\ $$$${ii}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {r}\left(_{{r}} ^{{n}} \right) \\ $$$${iii}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} \frac{\mathrm{1}}{{r}+\mathrm{1}}\left(_{{r}} ^{{n}} \right) \\ $$$${iv}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{2}{r}} ^{\mathrm{2}{n}} \right) \\ $$$${v}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} \left(_{{n}−{r}} ^{{n}+\mathrm{1}} \right) \\ $$$$\mathrm{3}.{Find}\:{the}\:{sum} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{r}=\mathrm{0}} {\overset{{n}−{k}} {\sum}}\left(_{{k}} ^{{n}−{r}} \right),\:\:\:{where}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…,{n} \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
1) let f(x)= 1+x +x^2 +...+x^n =Σ_(k=0) ^n x^k ⇒f^′ (x)=Σ_(k=1) ^n k x^(k−1) but if x≠1 f(x)= ((x^(n+1) −1)/(x−1)) ⇒f^′ (x)=((nx^(n+1) −(n+1)x^(n +1) +1)/((x−1)^2 )) so Σ_(k=1) ^n kx^(k−1) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) if x =1 f^′ (x)= 1+2+3+...+n ⇒f^′ (x)= ((n(n+1))/2) let take x=2 ⇒Σ_(k=1) ^n k.2^(k−1) =n 2^(n+1) −(n+1)2^n +1 Σ_(k=1) ^n k .2^k =n 2^(n+2) −(n+1)2^(n+1) +2 . =4n 2^n −(2n+2)2^n +2=(2n−2)2^n +2 =(n−1)2^(n+1) +2.
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)=\:\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \:+…+{x}^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \\ $$$$\Rightarrow{f}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:\:{but}\:{if}\:{x}\neq\mathrm{1} \\ $$$${f}\left({x}\right)=\:\frac{{x}^{{n}+\mathrm{1}} \:−\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow{f}^{'} \left({x}\right)=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}\:+\mathrm{1}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${so}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{kx}^{{k}−\mathrm{1}} \:=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$${if}\:{x}\:=\mathrm{1}\:\:\:{f}^{'} \left({x}\right)=\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}\: \\ $$$$\Rightarrow{f}^{'} \left({x}\right)=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\: \\ $$$${let}\:{take}\:{x}=\mathrm{2}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} {k}.\mathrm{2}^{{k}−\mathrm{1}} ={n}\:\mathrm{2}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}} \:+\mathrm{1} \\ $$$$\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:.\mathrm{2}^{{k}} \:={n}\:\mathrm{2}^{{n}+\mathrm{2}} \:−\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} \:+\mathrm{2}\:. \\ $$$$=\mathrm{4}{n}\:\mathrm{2}^{{n}} −\left(\mathrm{2}{n}+\mathrm{2}\right)\mathrm{2}^{{n}} \:+\mathrm{2}=\left(\mathrm{2}{n}−\mathrm{2}\right)\mathrm{2}^{{n}} \:+\mathrm{2} \\ $$$$=\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} \:+\mathrm{2}. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
2)ii let p(x)=Σ_(k=0) ^n C_n ^k x^k we have p(x)=(x+1)^n ⇒ Σ_(k=0) ^n (−1)^k C_n ^k =p(−1)=0
$$\left.\mathrm{2}\right){ii}\:{let}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{x}^{{k}} \:\:{we}\:{have} \\ $$$${p}\left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} \:\Rightarrow\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} ={p}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
iii let p(x)= Σ_(k=0) ^n C_n ^k x^k =(x+1)^n ⇒ ∫ p(x)dx =Σ_(k=0) ^n (1/(k+1)) C_n ^k x^(k+1) +c =(1/(n+1))(x+1)^(n+1) +c x=0 ⇒c=−(1/(n+1)) ⇒ Σ_(k=0) ^n (x^(k+1) /(k+1)) C_n ^k = (((x+1)^(n+1) −1)/(n+1)) let take x=−1 ⇒ −Σ_(k=0) ^n (((−1)^k )/(k+1)) C_n ^k =(((x+1)^(n+1) −1)/(n+1)) ⇒ Σ_(k=0) ^n (((−1)^k )/(k+1)) C_n ^k = ((1−(x+1)^(n+1) )/(n+1)) .
$${iii}\:{let}\:\:{p}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:=\left({x}+\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$$\int\:{p}\left({x}\right){dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:{x}^{{k}+\mathrm{1}} +{c}\:\:\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:+{c} \\ $$$${x}=\mathrm{0}\:\Rightarrow{c}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:=\:\frac{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\mathrm{1}}{{n}+\mathrm{1}}\:{let}\:{take}\:{x}=−\mathrm{1}\:\Rightarrow \\ $$$$ \\ $$$$−\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:\:=\frac{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:\:\:=\:\frac{\mathrm{1}−\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
⇒ Σ_(k=0) ^n (((−1)^k )/(k+1)) C_n ^k = (1/(n+1)) .
$$\Rightarrow\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \:\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:. \\ $$
Answered by ajfour last updated on 20/Jun/18
s_n =1+2x+3x^2 +4x^3 +....+nx^(n−1) xs_n = [x+2x^2 +3x^3 +....+(n−1)x^(n−1) +nx^n (1−x)s_n =1+x+x^2 +...+x^(n−1) −nx^n s_n =((1−x^n )/((1−x)^2 ))−((nx^n )/(1−x)) .
$${s}_{{n}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +….+{nx}^{{n}−\mathrm{1}} \\ $$$${xs}_{{n}} =\:\:\:\:\:\left[{x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +….+\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} +{nx}^{{n}} \right. \\ $$$$\left(\mathrm{1}−{x}\right){s}_{{n}} =\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}−\mathrm{1}} −{nx}^{{n}} \\ $$$${s}_{{n}} =\frac{\mathrm{1}−{x}^{{n}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{{nx}^{{n}} }{\mathrm{1}−{x}}\:. \\ $$

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