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Question Number 32295 by abdo imad last updated on 22/Mar/18
calculate Σ_(k=0) ^n (2k+1)(−1)^k .
$${calculate}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(\mathrm{2}{k}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{k}} \:\:. \\ $$
Commented by prof Abdo imad last updated on 04/Apr/18
let put S_n = Σ_(k=0) ^n (2k+1)(−1)^k =Σ_(k=2p) (...) + Σ_(k=2p+1) (...) = Σ_(p=0) ^([(n/2)]) (4p+1) −Σ_(p=0) ^([((n−1)/2)]) (4p+3) but the sequences (4p+1) and (4p+3) are arithmetic Σ_(p=0) ^([(n/2)]) (4p+1) = (([(n/2)] +1)/2)( 1 + 4[(n/2)] +1) =(1+[(n/2)])( 1+2[(n/2)]) Σ_(p=0) ^([((n−1)/2)]) (4p+3) = (([((n−1)/2)] +1)/2)( 3 +4[((n−1)/2)] +3 ) = (1 +[((n−1)/2)])( 3 +2[((n−1)/2)])
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{2}{k}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{2}{p}} \left(…\right)\:+\:\sum_{{k}=\mathrm{2}{p}+\mathrm{1}} \:\left(…\right) \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(\mathrm{4}{p}+\mathrm{1}\right)\:\:−\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\left(\mathrm{4}{p}+\mathrm{3}\right) \\ $$$${but}\:{the}\:{sequences}\:\left(\mathrm{4}{p}+\mathrm{1}\right)\:{and}\:\left(\mathrm{4}{p}+\mathrm{3}\right)\:{are}\:{arithmetic} \\ $$$$\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(\mathrm{4}{p}+\mathrm{1}\right)\:=\:\frac{\left[\frac{{n}}{\mathrm{2}}\right]\:+\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{1}\:+\:\mathrm{4}\left[\frac{{n}}{\mathrm{2}}\right]\:+\mathrm{1}\right) \\ $$$$=\left(\mathrm{1}+\left[\frac{{n}}{\mathrm{2}}\right]\right)\left(\:\mathrm{1}+\mathrm{2}\left[\frac{{n}}{\mathrm{2}}\right]\right) \\ $$$$\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\left(\mathrm{4}{p}+\mathrm{3}\right)\:=\:\frac{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\:+\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{3}\:\:+\mathrm{4}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\:+\mathrm{3}\:\right) \\ $$$$=\:\left(\mathrm{1}\:+\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\right)\left(\:\mathrm{3}\:\:+\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\right) \\ $$
Commented by prof Abdo imad last updated on 06/Apr/18
another method S_n = 2 Σ_(k=0) ^n k(−1)^k +Σ_(k=0) ^n (−1)^k but Σ_(k=0) ^n (−1)^k =((1−(−1)^(n+1) )/2) = ((1+(−1)^n )/2) Σ_(k=0) ^n k(−1)^k = −Σ_(k=1) ^n k(−1)^(k−1) for that let find Σ_(k=1) ^n k x^(k−1) we have for x≠1 Σ_(k=0) ^n x^k = ((x^(n+1) −1)/(x−1)) by derivation we get Σ_(k=1) ^n k x^(k−1) = ((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^n k (−1)^(k−1) = ((n(−1)^(n+1) −(n+1)(−1)^n +1)/4) = ((1−(2n+1)(−1)^n )/4) ⇒ S_n = ((1+(−1)^n )/2) + ((1−(2n+1)(−1)^n )/4) .
$${another}\:{method} \\ $$$${S}_{{n}} =\:\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\left(−\mathrm{1}\right)^{{k}} \:\:+\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}}\:=\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\left(−\mathrm{1}\right)^{{k}} \:=\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:{for}\:{that}\:{let} \\ $$$${find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{{k}} \:=\:\frac{{x}^{{n}+\mathrm{1}} \:−\mathrm{1}}{{x}−\mathrm{1}}\:\:{by}\:{derivation}\:{we}\:{get} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:=\:\frac{{nx}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:=\:\frac{{n}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{1}}{\mathrm{4}} \\ $$$$=\:\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}\:\Rightarrow \\ $$$${S}_{{n}} \:=\:\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\:\:+\:\frac{\mathrm{1}−\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}\:. \\ $$

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