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

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