Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 37890 by abdo mathsup 649 cc last updated on 19/Jun/18

calculate A_n = Σ_(k=0) ^n (−1)^k (2k+3) interms of n

$${calculate}\:{A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left(\mathrm{2}{k}+\mathrm{3}\right)\:{interms}\:{of}\:{n} \\ $$

Commented by prof Abdo imad last updated on 20/Jun/18

A_n =2 Σ_(k=0) ^n k(−1)^k +3 Σ_(k=0) ^n (−1)^k but Σ_(k=0) ^n (−1)^k = ((1−(−1)^(n+1) )/2) let p(x)=Σ_(k=0) ^n x^k we have p(x)=((x^(n+1) −1)/(x−1)) if x≠−1 and p^′ (x)=Σ_(k=1) ^n k x^(k−1) ⇒ x p^′ (x)=Σ_(k=1) ^n k x^k ⇒Σ_(k=1) ^n k(−1)^k =−p^′ (−1) p^′ (x)=(((n+1)x^n (x−1)−x^(n+1) +1)/((x−1)^2 )) =(((n+1)x^(n+1) −(n+1)x^n −x^(n+1) +1)/((x−1)^2 )) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ p^′ (−1)= ((n(−1)^(n+1) −(n+1)(−1)^n +1)/4) =((−n(−1)^n −(n+1)(−1)^n +1)/4) =((1−(2n+1)(−1)^n )/4) ⇒ A_n = 2 (((2n+1)(−1)^n −1)/4) +(3/2)( 1+(−1)^n ) = (((2n+1)(−1)^n −1 +3 +3(−1)^n )/2) =(((2n+4)(−1)^n +2)/2) A_n =(n+2)(−1)^n +1 .

$${A}_{{n}} =\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\left(−\mathrm{1}\right)^{{k}} \:+\mathrm{3}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:=\:\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}} \\ $$$${let}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:\:{we}\:{have}\:{p}\left({x}\right)=\frac{{x}^{{n}+\mathrm{1}} \:−\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${if}\:{x}\neq−\mathrm{1}\:\:{and}\:{p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${x}\:{p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}} \:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left(−\mathrm{1}\right)^{{k}} =−{p}^{'} \left(−\mathrm{1}\right) \\ $$$${p}^{'} \left({x}\right)=\frac{\left({n}+\mathrm{1}\right){x}^{{n}} \left({x}−\mathrm{1}\right)−{x}^{{n}+\mathrm{1}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){x}^{{n}} \:−{x}^{{n}+\mathrm{1}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{nx}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${p}^{'} \left(−\mathrm{1}\right)=\:\frac{{n}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{−{n}\left(−\mathrm{1}\right)^{{n}} −\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\mathrm{4}}\:\Rightarrow \\ $$$${A}_{{n}} \:=\:\mathrm{2}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{2}}\left(\:\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right) \\ $$$$=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\:+\mathrm{3}\:+\mathrm{3}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{4}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{2}}{\mathrm{2}} \\ $$$${A}_{{n}} =\left({n}+\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}\:. \\ $$

Commented by math khazana by abdo last updated on 20/Jun/18

x≠1

$${x}\neq\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com